ENGINEERING  LIBRARY 


STEAM  CHARTS 


ALSO 


A  Table  of  Theoretical  Jet  Velocities  and  The 
Corrections  of  Mercury  Columns 


WITH 


FIFTY  ILLUSTRATIVE  PROBLEMS 


BY 

F.   O.   ELLENWOOD 

Assistant  Professor  of  Heat  Power  Engineering,  Cornell  University 
Member  of  the  American  Society  of  Mechanical  Engineers 


FIRST  EDITION 
FIRST    THOUSAND 


NEW  YORK 

JOHN  WILEY  &  SONS,  INC. 

LONDON:    CHAPMAN    &    HALL,    LIMITED 
1914 


Copyright,  1914,  by 
F.    O.    ELLENWOOD 

Copyrighted  in  Great  Britain 


ENGINEERING  LIBRARY 


PUBLISHERS  PRINTING  COMPANY 
»07-2i7  Wett  Twenty-fifth  Street,  New  Yor* 


PREFACE 

THIS  little  book  is  intended  to  be  of  assistance  to  engineers  and 
students  when  making  calculations  involving  wet  or  superheated 
steam.  The  chief  aim  of  the  author  has  been  to  prepare  a  set  of 
steam  charts  which  shall  be  accurate  and  comprehensive,  and  at 
the  same  time  convenient  to  handle,  and  easy  to  read.  An  attempt 
has  also  been  made  to  give,  concisely,  the  corrections  to  be  applied 
to  the  readings  of  mercury  columns,  and  to  prepare  a  table  of 
velocities,  which  it  is  hoped  may  prove  useful. 

In  order  to  illustrate  some  of  the  uses  of  the  charts  and  tables, 
and  also  to  aid  those  who  may  desire  it,  a  number  of  problems,  with 
their  solutions,  have  been  added.  To  make  these  of  more  assistance, 
they  have  been  indexed.  For  the  further  aid  of  those  who  may 
desire  a  brief  review  of  the  thermodynamics  of  steam,  and  in  order 
to  make  clear  the  meaning  of  all  terms  used,  the  few  pages  of 
Fundamental  Principles  were  written. 

For  the  main  chart,  total  heat  and  specific  volume  were  chosen 
as  coordinates  because  of  the  fact  that  upon  these  two  values 
could  be  plotted  lines  of  constant  pressure,  entropy  and  quality  (or 
superheat),  so  that  each  pair  of  the  five  sets  of  lines  will  make  clear 
intersections.  The  total  heat  entropy  chart  does  not  permit  this. 
To  complete  the  set  of  values  ordinarily  needed,  the  curve  was 
added,  showing  the  heat  of  the  liquid  and  temperature  of  vapori- 
zation. The  supplementary  chart,  Plate  8,  enables  one  to  read  the 
external  work,  and  therefore  obtain  easily  the  intrinsic  heat.  The 
index  chart  for  Plates  1  to  6  was  made  to  give  a  general  idea  of  the 
relative  position  and  shape  of  each  set  of  lines,  to  show  quickly 
the  limiting  values  for  each  of  the  six  sections,  and  to  assist  in 
determining  the  particular  plate  needed. 

The  range  of  pressures,  qualities,  and  superheats  is  intended  to 
be  more  than  sufficient  for  present  practice.  For  the  wet  region 
the  inch  of  mercury  was  used  as  the  main  unit  to  represent  pressures 

iii 


IV  PREFACE 

less  than  one  pound  absolute,  as  it  is  believed  that  this  is  the  more 
convenient  one  for  practical  work.  Special  endeavor  has  been 
made  to  prevent  confusion  of  these  two  units  by  using  broken  lines 
to  represent  pressures  hi  inches  of  mercury,  and  by  putting  the 
proper  units  with  each  numeral  representing  pressure  in  this  region. 

The  book  form  of  chart  was  chosen  because  the  author  believes 
that  it  will  be  of  greater  convenience  and  easier  to  read  than  a  large 
folded  chart  made  to  the  same  scales.  By  making  the  plates  small 
the  eye  has  to  travel  only  a  short  distance  to  read  the  scales,  and 
this  may  also  be  done  without  requiring  any  desk  space  whatever. 
The  book  form  also  has  the  advantages  of  better  protecting  the 
chart,  permitting  a  quicker  reference,  and  wasting  less  space  in  the 
corners,  than  does  the  same  chart  when  hi  the  form  of  a  large 
folded  sheet. 

To  Prof.  Lionel  S.  Marks  and  to  Dr.  Harvey  N.  Davis,  and  to 
their  publishers,  Longmans,  Green  &  Co.,  the  author  desires  to 
express  his  thanks  for  permission  to  use  then-  steam  tables  hi  pre- 
paring these  charts.  He  also  wishes  to  acknowledge  his  indebted- 
ness to  Prof.  Albert  W.  Smith,  Director  of  Sibley  College,  and  to 
Prof.  William  N.  Barnard,  for  then*  many  helpful  criticisms;  and 
to  Mr.  C.  H.  Berry  and  Mr.  E.  T.  Jones,  instructors  in  Sibley 
College,  for  their  able  assistance  in  preparing  the  charts  and 
problems.  F.  0.  E. 

ITHACA,  New  York,  August,  1914. 


CONTENTS 

PAGE 

INTRODUCTION 

Fundamental  Principles 1 

Preparation  and  Use  of  the  Steam  Charts  and  Tables      ....  14 

Atmospheric  Pressure  and  Barometric  Corrections       18 

CHARTS 

Index  Chart       ..................  23 

Plates  1  to  7,  the  Total  Heat-Volume  Chart 24 

Plates  8a  and  8b,  the  External  Work-Volume  Chart 38 

Plate  9a,  Correction  of  Mercury  Column  Due  to  Temperature  .     .  40 
Plate  9b,  Correction  of  Barometric  Readings  Due  to  Change  in 
Elevation    .     .     * 40 

TABLES 

1.  Correction  of  Barometric  Readings  to  45°  Latitude     ....  41 

2.  Correction  of  the  Barometer  for  Capillarity 41 

3.  Density  of  Mercury 41 

4.  Theoretical  Velocities  of  Steam 42 

PROBLEMS   .    ....    . . 45 

INDEX .87 


INTRODUCTION 

FUNDAMENTAL  PRINCIPLES 

Pressure- Volume  and  Temperature-Entropy  Diagrams. — In  the 

study  of  thermodynamics  of  vapors  the  pressure-volume  and  the 
temperature-entropy  diagrams  are  of  very  great  importance,  for  the 
reason  that  by  then-  aid  the  areas  representing  work  and  heat, 
respectively,  may  usually  be  shown.  Since  the  engineer  is  con- 
cerned with  the  transformation  of  heat  into  work,  any  diagrams 
which  will  assist  him  to  understand  how  this  is  accomplished  will 
always  be  very  useful.  Other  diagrams  may  be  of  more  assistance 
in  obtaining  numerical  results,  but  these  two  will  always  be  fore- 
most in  analyzing  thermodynamic  processes. 

Pressure- Volume  Diagram. — Referring  to  Fig.  la,  where  the 
absolute  pressure,  P,  in  pounds  per  square  foot,  is  represented  by 
the  ordinates,  and  the  volume,  V,  in  cubic  feet  by  abscissae,  the 
area  abhg  underneath  curve  ab  represents  the  work  done  in  foot- 
pounds by  the  substance  in  passing  along  the  constant-pressure  line 
from  a  to  b.  This  might  be  expressed  in  this  manner: 

Work~|6  =  f^rb  PdV  =  P  [V&  -  VJ  =  area  abhg. 

-*a       J    \a 

In  general,  it  may  be  stated  that  if  a  substance  expands  or  is 
compressed  in  such  a  manner  that  its  pressure-volume  history  is 
definite  for  the  entire  change,  as  from  the  point  a  to  some  point  c 
along  the  path  adc,  Fig.  la,  the  following  equation  is  true: 

WorkT  =   /V  PdV  =  area  adchg. 

Ja        *    Va 

This  would  represent  work  done  by  the  substance  upon  some 
other  body.  In  case  the  substance  had  been  compressed  along  the 

1 


FUNDAMENTAL   PRINCIPLES 


path  cda,  then  the  area  cdagh  represents  the  work  done  upon  this 
substance  in  order  to  compress  it, 

-.a          /»V  rV 

or  Work  1    =   /    "  PdV  =  -   /  __c  PdV 

Jc        J    \c  J    Va 

=  —  [area  adchg]  =  area  cdagh. 

Temperature-Entropy  Diagram. — If,  by  the  addition  of  heat,  a 
substance  may  be  made  to  change  its  state  point  a  to  some  other 


Volume 

(a) 


Entropy 

(b) 


FIG.  1. 


condition  6,  in  such  a  manner  that  its  temperature  and  pressure 

are  uniform*  throughout  the  entire  mass  of  the  substance  during 

this  change,  then  the  heat  which  has  been  added  to  this  substance 

is  equal  to  the  area  mabn,  Fig.  Ib.    This  is  true  by  reason  of  the 

definition  of  entropy.    In  other  words,  for  such  processes  as  this,  if 

H  =  Heat  added  to  the  substance  in  B.  t.  u. 

T  =  Absolute  temperature  (=  460  +  t°  F.) 

<t>  =  Entropy 

then  the  change  of  entropy  from  a  to  b  is  defined  by  the  equation 
,.       dH 


or 


I6 

Hl  = 


=  area  abnm 


*  It  is  particularly  important  not  to  confuse  this  word  uniform  with  the  word  constant. 


FUNDAMENTAL   PRINCIPLES 


For  the  condition  given  above,  if  the  heat  had  been  added  at 
constant  temperature,  as  from  b  to  /,  Fig.  Ib,  then, 


0  =  T    0/-  —  0&      =  area  bfkn 

or,  0y  —  05  =  Tp~  =  distance  &/  or  nfc. 

For  both  of  these  cases  the  addition  of  heat  to  the  substance  caused 
an  increase  of  entropy.  Had  the  heat  been  abstracted,  the  entropy 
would  have  been  decreased.  Note  that  it  is  always  the  change  in 
entropy  that  engineers  are  interested  in,  rather  than  its  absolute 
values. 

Adiabatics. — The  term  adiabatic  means  no  transfer  of  heat. 
Hence  an  adiabatic  expansion  or  compression  means  one  in  which 
no  heat  is  added  or  abstracted  during  the  process. 

Reversible  and  Irreversible  Adiabatics. — The  two  general 
classes  into  which  all  adiabatics  may  be  divided,  are  called  reversible 
and  irreversible  adiabatics.  An  adiabatic  expansion  or  compres- 
sion, which  is  frictionless  and  which  takes  place  in  such  a  manner 
that  the  substance  passes  through  a  continuous  series  of  uniform 
states,  would  be  called  a  reversible  adiabatic.  A  reversible  adia- 
batic is  also  called  a  constant  entropy  line  or  isentropic,  because  a 
vertical  line  on  the  temperature-entropy  chart  is  the  only  one  which 
permits  the.  area  underneath  it  to  become  equal  to  zero,  thus  satis- 
fying the  definitions  of  adiabatics  and  entropy. 

It  is  impossible  to  have  these  conditions  fulfilled  in  actual  cases, 
but  it  is  us'eful  to  study  them  and  compare  them  with  the  actual 
conditions  which  may  be  made  to  approach  very  close  to  the  ideal. 
Thus,  a  substance  may  expand  in  a  single  cylinder,  with  but  very 
little  friction,  and  with  but  very  little  transfer  of  heat  to  or  from 
the  cylinder  walls,  and  slowly  enough  so  that  all  of  the  substance 
in  this  cylinder  is  almost  exactly  in  the  same  state.  Such  cases 
may  properly  be  treated  as  reversible  adiabatics  or  isentropics. 

On  the  other  hand,  sudden  or  "free"  expansions  are  sure  to 
set  up  eddies  whose  kinetic  energy  will  soon  reappear  in  the  form 
of  heat,  thus  causing  an  increase  of  entropy.  Friction  will  have  a 


4  FUNDAMENTAL   PRINCIPLES 

similar  effect.  Such  expansions  are  called  irreversible,  and  if  they 
have  taken  place  without  any  heat  being  transferred  to  or  from 
another  substance,  they  would  be  called  irreversible  adiabatics. 

In  the  case  of  adiabatic  expansion  of  gases  or  steam  through  a 
properly-formed  nozzle  in  which  there  are  no  eddies  formed,  and  in 
which  there  is  no  friction,  the  transformation  of  the  available  heat 
energy  into  velocity  is  complete  and  may  therefore  be  considered  as 
a  reversible  adiabatic.  In  actual  steam  turbine  nozzles  the  loss  due 
to  friction  and  eddies  is  extremely  small,  so  that  in  turbine  design 
the  expansion  in  the  nozzle  is  justly  assumed  to  be  a  reversible 
adiabatic.  The  main  losses  in  the  turbine  occur  in  the  endeavor 
to  transfer  the  energy  of  the  jet  to  the  turbine  blades.  These  losses 
cause  considerable  reheating,  so  that  in  any  actual  case  there  is 
considerable  increase  in  the  entropy  of  the  steam  in  passing  through 
the  turbine.  As  an  illustration,  see  problems  41  and  42. 

The  term  adiabatic  will  be  used  hereafter  to  mean  reversible 
adiabatic,  unless  specifically  stated  otherwise. 

Work  and  Heat  Depend  upon  Path. — From  the  consideration  of 
the  PV  and  T<£  charts  it  will  be  evident  that  it  is  not  sufficient  to 
give  merely  the  initial  and  final  states  of  a  substance  in  order  to 
find  the  work  done  and  the  heat  required  to  go  from  one  state  to 
another.  It  is  also  necessary  to  give  the  exact  path  to  be  followed. 
Thus,  in  Figs,  la  and  Ib,  heat  might  be  added  and  abstracted  in 
such  a  manner  that  the  state  c  is  finally  reached  by  means  of  the 
constant  pressure  line  a&,  and  the  constant  volume  line  be,  or 
this  same  state  might  have  been  reached  more  directly  by  some 
such  line  as  adc. 

By  observing  the  areas  it  will  be  seen  that  the  work  done  in  the 
first  case  is  abhg,  while  for  the  path  adc  the  work  done  is  represented 
by  the  area  adchg.  Also  from  the  T<£  chart  it  may  be  seen  that  in 
going  from  a  to  6  an  amount  of  heat  equal  to  the  area  abnm  was 
supplied  to  the  substance,  and  in  going  along  the  constant  volume 
line,  be,  an  amount  of  heat  equal  to  the  area  bcmn  was  abstracted 
from  the  substance.  The  substance  might  also  have  reached  the 
state  c  by  expanding  adiabatically  along  the  line  adc;  but  note 
the  difference  in  areas  representing  the  work  and  heat. 


FUNDAMENTAL   PRINCIPLES  5 

Cycles. — If  a  substance  should  be  made  to  follow  a  series  of 
paths  such  as  ab,  be,  and  ca,  so  that  it  is  returned  to  its  initial 
state,  the  substance  would  then  have  completed  a  cycle,  and  the 
substance  used  would  have  been  called  the  working  substance. 
Had  the  cycle  been  completed  in  the  order  of  the  letters,  dbc,  an 
amount  of  work  in  foot-pounds,  equal  to  the  area,  a&c,  Fig.  la, 
would  have  been  done  upon  some  external  mechanism.  The  heat 
equivalent  in  B.  t.  u.  of  this  work,  would  be  represented  by  the 
area  abc,  Fig.  Ib.  This  heat  would  have  been  supplied  to  the 
working  substance  from  some  external  source. 

Properties  which  are  Independent  of  the  Path. — After  a  cycle 
has  been  completed,  the  substance  is  in  exactly  the  same  condition 
as  at  the  beginning,  so  its  pressure,  volume,  temperature,  entropy, 
and  intrinsic  heat  must  all  be  the  same  as  originally.  Likewise,  if  a 
substance  is  in  a  certain  state,  such  as  c,  it  merely  means  that  all 
of  the  above  properties  have  some  definite  value  and  may  be  determined 
regardless  of  the  manner  in  which  this  substance  may  have  reached 
this  state  c. 

Intrinsic  Heat  or  Intrinsic  Energy. — Either  of  these  terms  may 
be  used  to  represent  all  of  the  heat  energy  contained  within  a 
substance  measured  above  some  convenient  standard.  It  is  to  be 
carefully  noted  that  this  is  a  different  quantity  in  general  from  the 
heat  required  to  bring  the  substance  to  a  particular  state  from  the 
standard  condition.  This  is  on  account  of  the  fact  that,  when  heat 
is  added  to  a  substance,  in  which  no  heat  is  lost  by  radiation  or 
otherwise,  in  which  there  is  no  change  in  electrical  or  chemical 
energy,  or  in  which  there  is  no  change  in  the  kinetic  energy  of  the 
substance  due  to  its  mass  velocity,  then  in  going  from  a  to  b,  a 
general  equation  may  be  written,  thus: 

Heat~|6  _  Gain  in  intrinsic-]6    ,    External  work"]6 
added  J  a  heat  Ja  done         Ja 

The  external  work  done  means  that  work  is  done  by  the  sub- 
stance upon  some  external  mechanism.  In  case  this  term  is  nega- 
tive, it  means  that  the  work  is  done  by  the  external  mechanism 
upon  the  substance. 


FUNDAMENTAL   PRINCIPLES 


Steam  Formed  at  Constant  Pressure. — For  nearly  all  commercial 
purposes  steam  is  generated  in  a  steam  boiler  at  a  pressure  which 
is  maintained  nearly  constant.  It  is  therefore  very  important  to 
consider  carefully  the  formation  of  steam  at  constant  pressure.  It 
has  been  found  by  experiment  that  during  the  change  of  state  of 
any  substance  from  a  liquid  to  a  vapor,  that  if  the  pressure  be  kept 
constant  the  temperature  of  the  vapor  in  contact  with  its  liquid  will 
also  remain  constant  until  the  vaporization  has  been  completed. 


Ul 


n' 


\n  n 


Volume 

(a) 


Entropy 


FIG.  2. 


Suppose  that  a  pound  of  water  at  a  temperature  of  melting  ice, 
32°  F.,  is  contained  in  a  metal  cylinder  with  a  tight-fitting  piston 
resting  upon  the  surface  of  the  water.  The  area  of  this  piston  and 
the  total  weight  resting  upon  the  water  being  known,  the  abso- 
lute pressure  on  the  water  may  be  determined.  Let  this  value  hi 
pounds  per  square  inch  be  denoted  by  p,  and  in  pounds  per  square 
foot  by  P.  The  volume  of  this  pound  of  water  will  be  very  small, 
only  .016  cubic  feet.  We  may  represent  this  starting  point  by  the 
letter  a  in  Figs.  2a  and  2b.  (We  locate  the  point  a  on  the  zero 
entropy  line  merely  as  a  matter  of  convenience,  as  we  shall  be  con- 
cerned only  with  the  heat  measured  above  32°  F.) 


FUNDAMENTAL   PRINCIPLES  7 

Let  heat  be  applied  to  the  cylinder  now,  and  it  will  be  found  that 
at  first  the  piston,  moves  up  only  a  little,  due  to  the  slight  increase 
in  the  volume  of  the  water.  The  temperature,  however,  is  ob- 
served to  rise  rapidly  until  it  reaches  some  definite  point  dependent 
entirely  upon  the  pressure  imposed  by  the  piston.  Let  this  state 
be  represented  by  b.  The  substance  is  now  all  in  the  form  of 
liquid,  but  at  the  particular  temperature  known  as  the  temperature 
of  vaporization  for  this  given  pressure. 

Since  the  volume  has  changed  only  very  slightly,  the  points  a 
and  b  on  the  PV  diagram  will  almost  exactly  coincide.  Even 
though  this  water  had  been  heated  to  a  temperature  of  400°  F.,  as 
it  might  have  been  if  under  a  pressure  of  250  pounds  or  more,  its 
volume  would  still  only  be  .0187  cubic  foot.  This  extremely  small 
change  in  volume  means  that  very  little  work,  shown  by  the  area 
abkk',  Fig.  2a,  has  been  done  on  the  piston,  and  consequently  all 
the  heat  that  has  been  applied  has  been  used  to  increase  the  sen- 
sible heat  of  tjie  water. 

The  Heat  of  the  Liquid  is  the  term  applied  to  this  heat,  which 
is  used  to  heat  a  unit  weight  of  water  from  32°  F.  to  the  temperature 
of  vaporization.  The  area  representing  the  heat  of  the  liquid  is 
abmim,  Fig.  2b.  With  further  addition  of  heat  it  will  now  be 
found  that  the  volume  is  increased  very  much  and  that  the  temper- 
ature remains  constant  until  some  state  c  is  reached.  The  water 
has  now  all  been  evaporated,  and  exists  in  the  form  of  dry  saturated 
steam.  Any  further  addition  of  heat  will  cause  it  to  become  super- 
heated, and  any  abstraction  of  heat  will  cause  it  to  be  partially 
condensed.  The  heat  which  was  necessary  to  vaporize  this  pound 
of  water  under  this  pressure  is  represented  by  the  area  be  n  Wi,  and 
is  called  the  latent  heat  of  vaporization,  or  latent  heat  of  steam. 

The  Total  Heat  of  Dry  Saturated  Steam  is  equal  to  the  heat  of 
the  liquid  plus  the  latent  heat  of  vaporization  and  is  equal  to  the 
area  abcnm,  Fig.  2b. 

The  Saturation  Curve  on  any  diagram  is  the  curve  which  shows 
the  two  coordinate  values  of  dry  saturated  steam  for  the  entire 
range  of  scales  on  that  diagram.  Thus  in  Fig.  2a  the  specific  volume 
of  dry  saturated  steam  may  be  determined  for  any  pressure  by 


8  FUNDAMENTAL   PRINCIPLES 

means  of  the  saturation  curve,  c  c\  c2.    The  saturation  curve  may 
also  be  seen  for  other  coordinates  as  in  Fig.  2b,  Plate  Ib,  or  Plate  8a. 

Wet  Steam  and  Quality. — During  the  formation  of  a  pound  of 
dry  saturated  steam  in  the  manner  just  outlined,  it  might  have 
been  possible  to  stop  the  addition  of  heat  before  all  the  water  was 
evaporated  and  the  resultant  mixture  of  water  and  steam  would  be 
what  is  known  as  wet  steam.  That  portion  by  weight  of  this  mix- 
ture which  is  dry  vapor  would  be  known  as  the  Quality  of  the 
steam.  Thus  if  x  represents  quality  and  if  nine-tenths  of  the 
pound  of  water  have  been  vaporized,  this  fact  would  be  expressed 
by  the  equation  x  =  90%.  In  Fig.  2b  the  quality  at  c  is  unity  and 
at  6  the  quality  is  zero;  or  Xc  =  100%  and  x&  =  0.  It  is  im- 
portant to  note  that  this  interpretation  of  quality  is  the  most 
useful  one  in  using  the  T<£  chart.  In  this  chart  it  is  understood  that 
a  unit  weight  of  substance  is  being  considered  unless  it  be  defi- 
nitely stated  otherwise,  and  it  is  sometimes  necessary  to  carry  this 
unit  weight  of  substance  throughout  the  extreme  range  of  quality 
in  order  to  analyze  a  cycle  or  to  understand  many  important 
thermodynamic  relations. 

On  the  other  hand,  steam  coming  from  a  boiler  often  carries 
with  it  a  considerable  amount  of  moisture.  If  a  sample  of  this 
mixture  be  obtained,  then  the  weight  of  the  dry  steam  present 
divided  by  the  weight  of  this  entire  mixture  would  be  the  quality 
of  the  steam  coming  from  the  boiler,  the  weight  of  water  still  re- 
maining in  the  boiler  not  being  considered  at  all  in  this  determi- 
nation of  quality. 

The  Total  Heat  of  Wet  Steam  is  equal  to  the  heat  of  the  liquid 
plus  the  quality  multiplied  by  the  latent  heat  of  vaporization,  or 
h  +  xL,  where  h  represents  the  heat  of  the  liquid  and  L  the  latent 
heat.  If  c',  Fig.  2b,  represents  this  state  of  the  steam,  then  the 
area  dbc'n'm  is  equal  to  the  total  heat  at  c'. 

Superheated  Steam. — If  heat  is  added  to  dry  saturated  steam, 
the  pressure  remaining  constant  as  before,  the  volume  and  temper- 
ature will  be  found  to  increase;  and  the  steam  is  said  to  be  super- 
heated. Superheated  steam  may  therefore  be  defined  as  any 
steam,  regardless  of  how  formed,  having  a  temperature  higher  than 


FUNDAMENTAL   PRINCIPLES  9 

the  temperature  of  saturated  steam  of  the  same  pressure.  The 
difference  between  these  two  temperatures  is  called  the  Degrees  of 
superheat,  and  is  nearly  always  represented  by  the  letter  D.  For 
the  state  d,  Fig.  2b,  D  =  Td  —  Tc.  If  the  specific  heat  of  super- 
heated steam  at  constant  pressure  be  represented  by  cp,  then  cp  D 
is  equal  to  the  area  cdnin  and  is  called  by  either  of  the  terms,  heat 
of  superheat,  heat  of  superheating  or  superheat. 

The  value  of  the  specific  heat  of  superheat  steam  is  variable, 
and  depends  upon  both  the  pressure  and  the  degrees  of  superheat. 

The  Total  Heat  of  Superheated  Steam  is  equal  to  the  total  heat 
of  dry  saturated  steam  plus  the  heat  of  superheat.  Thus  for  the 
state  point  d,  Fig.  2b,  the  total  heat  at  d  is  equal  to  the  area  abcdnim. 

Total  Heat  may  now  most  conveniently  be  defined  in  the  manner 
in  which  it  is  used  by  engineers,  that  is,  a  general  definition  which 
will  hold  for  wet,  dry,  or  superheated  steam. 

Total  Heat  of  Steam  in  any  Given  State  is  the  amount  of  heat 
required  to  heat  at  constant  pressure  a  unit  weight  of  water  from 
the  temperature  of  melting  ice  to  the  state  under  consideration. 

Thus  referring  to  Fig.  2b : 

Total  Heat  I    =  area  dbc'rim  =  h  +  xc'  L 

Total  Heat  I    =  area  abc  n  m  =  h  +  L 

Total  Heat  I    =  area  abcdnim  =  h  +  L  +  cpD. 

-Jd 

In  the  United  States  the  units  are  almost  always  B.  t.  u.  per 
pound.  All  values  of  total  heat  which  are  likely  to  be  needed  will 
be  found  on  Plates  1  to  7  inclusive. 

External  Work  at  Constant  Pressure. — During  the  period  of 
vaporization  the  volume  of  a  pound  of  water  is  changed  to  the  very 
much  larger  volume  of  dry  saturated  steam,  as  from  V&  to  Vc, 
Fig.  2a. 

If  this  volume  is  represented  in  cubic  feet  by  V  and  the  pressure 


10  FUNDAMENTAL   PRINCIPLES 

in  pounds  per  square  foot  by  P,  the  work  done  during  vaporization  is 
Work]^  =  P  [vc  -  V&]  ft.-lbs.  =  wezbchk, 

or  =  AP  (Vc  -  V&)  B.  t.  u. 
A  being  taken  as  the  reciprocal  of  the  mechanical  equivalent  of 

heat,  or  A  =        or 


AP  (Vc  —  Vfc)  is  commonly  known  as  the  external  latent  heat  of 
vaporization. 

During  superheating  the  external  work  done  is 


Work        =  PVd-Vc      ft.-lbs.  =  area  cdhji 
or  =  AP  (Vd  -  Vc)  B.  t.  u. 


For  the  entire  process  starting  with  water  at  32°  when  its 
specific  volume  would  accordingly  be  .016  cubic  feet  per  pound, 
the  external  work  done  in  order  to  reach  some  state  such  as  c'  or  d 
would  be 

Work]^  =  P  [vc'  —  Va]  ft.-lbs.  =  AP  [Vc'  -  .016  ]  B.  t.  u. 

Work]**  =  P  [Vd  -  Va  ]  ft.-lbs.  =  AP  [vd  -  .016]  B.  t.  u. 

This  is  what  is  called  the  constant  pressure  external  work,  and  is 
given  for  all  conditions  of  steam  by  Plates  8a  and  8b. 

This  work  has  been  done  upon  the  piston,  so  that  after  some 
state  such  as  d  has  been  reached,  the  piston  and  all  weights  resting 
upon  it  have  more  potential  energy  by  the  amount  of  AP  [  Vd  —  VJ 
B.  t.  u.  than  they  had  at  the  beginning  of  application  of  heat 
to  the  water  in  condition  a.  This  increase  in  potential  energy  of 
the  piston  and  its  weights  has  come  from  the  heat  which  was  neces- 
sary to  form  the  pound  of  steam  along  the  constant  pressure  path 
abed.  Now,  if  from  the  total  amount  of  heat  which  has  been 
added  to  a  substance  in  order  to  reach  a  certain  state  by  going 
along  a  certain  path,  all  of  the  heat  which  has  been  used  to  do 
external  work  be  subtracted,  the  amount  left  would  be  the  gain  in 
intrinsic  energy  of  the  substance,  provided  there  were  no  losses 


FUNDAMENTAL   PRINCIPLES  11 

and  that  no  energy  had  been  used  to  impart  velocity  to  the  sub- 
stance. 

The  Intrinsic  Energy,  Internal  Energy,  or  Intrinsic  Heat  of 

steam,  then,  merely  means  the  heat  energy  contained  within  the 
steam  above  32°,  and  may  always  be  found  by  subtracting  the 
constant  pressure  external  work  from  the  total  heat.  The  total 
heat  may  be  obtained  directly  from  Plates  1  to  7,  while  the  external 
work  may  be  obtained  from  Plates  8a  and  8b.  Since  we  can  not 
reduce  the  temperature  of  any  substance  to  absolute  zero,  it  is  not 
possible  to  reach  the  state  in  which  the  intrinsic  heat  is  zero.  For 
work  with  steam  we  need  only  be  concerned  with  the  intrinsic  heat 
above  32°  F. 

Specific  Volume  of  steam  means  the  volume  of  one  pound  of 
the  steam  in  its  given  state.  Unless  some  quality  or  superheat  is 
given,  it  is  understood  to  refer  to  dry  saturated  steam. 

The  specific  volume  of  wet  steam  may  be  determined  by  means 
of  the  quality  in  the  following  manner : 

Let  Vsat.  =  volume  of  1  Ib.  of  dry  saturated  steam  of  given  pres- 
sure, 

Vu;  =  volume  of  1  Ib.  of  water  at  the  temperature  of  vapori- 
zation, 

=  .016  cu.  ft.  at  32°  F., 
=  .017  cu.  ft.  at  250°  F., 
=  .018  cu.  ft.  at  350°  F., 
=  .019  cu.  ft.  at  420°  F. 
Let  u  =  change  of  volume  during  vaporization, 

=  Vsat.  -  Vw 

Then  for  some  point  such  as  c'  having  a  quality  xc',  the  specific 
volume  is 

Vc'  =  Vw  +  xc'u, 

=  Vw  +  Xc'  (Vsat.  ~  Vw), 
=  (1  -  XC')  Vw  +  XC'  VSat. 

This  becomes  equal  to  xc'  Vsat.  almost  exactly,  except  when  the 
quality  is  very  low  or  the  pressure  is  higher  than  atmospheric. 


12  FUNDAMENTAL   PRINCIPLES 

The  specific  volume  may  be  read  directly  from  Plates  1  to  8  inclu- 
sive. 

Entropies.  —  For  convenience,  entropy  of  water  at  32°  is  taken  as 
zero.  Hence  the  entropy  of  the  liquid  per  pound  of  water,  from 
Fig.  2b,  is 

dH        Aft  C  dT 


Where  cp  =  the  specific  heat  of  water  at  constant  pressure.  This 
value  depends  upon  the  temperature,  so  the  integration  is  not 
often  made,  but  instead  the  entropy  of  the  liquid  is  usually  obtained 
from  the  steam  tables. 

The  Entropy  of  Vaporization  is,  from  Fig.  2b, 

rcdH_   JL^      Latent  heat 
0c  -  06  -  Jb    TJT   =T^~abs.  Temp. 

The  Entropy  of  Superheat  or  the  change  of  entropy  due  to 
superheating  is,  from  Fig.  2b, 


_ 

<t>d    -  4>C    -=    Jc       -rf      = 

where  cp  represents  the  constant  pressure  specific  heat  of  super- 
heated steam. 

The  Entropy  of  Steam  in  any  state  means  the  total  entropy  up 
to  that  state,  measured  above  the  assumed  zero  of  entropy. 

For  Wet  Steam  the  Total  Entropy  is  equal  to 
Entropy  of  the  liquid  +  (Quality)  (Entropy  of  vaporization) 

For  Superheated  Steam  the  Total  Entropy  is  equal  to 
Entropy  of  the  liquid  +  Entropy  of  vaporization  +  Entropy  of 
Superheat 

The  Total  Entropies  are  plotted  on  Plates  1  to  7  inclusive  and 
these  should  serve  nearly  every  purpose  for  which  the  engineer 
has  need  of  their  numerical  values. 

Rankine  and  Clausius  Cycles.  —  The  most  useful  cycle  in  power- 
plant  work  is  the  one  represented  by  dbfe,  Figs.  5  and  6,  Problem  29. 
This  cycle  is  sometimes  known  as  the  Clausius  Cycle  and  some- 
times as  the  Rankine  Cycle,  but  since  the  analysis  of  this  cycle  was 


FUNDAMENTAL   PRINCIPLES  13 

first  published  by  each  of  these  men  at  the  same  time,  there  will 
probably  always  be  a  difference  of  opinion  as  to  which  name  should 
be  used. 

If,  instead  of  allowing  the  adiabatic  expansion  to  continue  until 
the  back  pressure  is  reached,  as  at  /,  the  expansion  had  been 
stopped  at  the  point  c,  thus  cutting  off  the  "toe"  in  each  diagram, 
there  would  be  another  cycle,  abode,  which  is  preferred  by  many 
engineers  as  a  basis  of  comparison  for  the  performance  of  recipro- 
cating engines.  This  latter  cycle  is  oftentimes  called  the  Rankine 
and  the  other  the  Clausius,  thus  making  a  convenient  distinction. 
They  are  also  spoken  of  as  the  "complete  expansion"  and  the 
"incomplete  expansion"  cycles.  These  latter  terms  will  be  used  in 
the  problems  of  this  book  in  order  that  there  may  be  no  possible 
confusion  of  the  two  cycles  by  using  the  terms  Rankine  and 
Clausius. 

Available  Energy  is  a  term  that  may  have  many  meanings,  but 
when  used  in  connection  with  steam  cycles  it  means  the  energy  of 
the  steam  which  would  be  converted  into  work  by  an  ideal  mech- 
anism that  would  carry  out  exactly  the  theoretical  cycle.  Thus, 
in  using  Table  IV,  the  engineer  would  consider  the  term  to  represent 
the  net  area  of  the  complete  expansion  cycle,  by  assuming  that  all 
of  this  energy  is  available  for  the  production  of  velocity  hi  the 
ideal  nozzle.  Then  with  an  ideal  turbine  all  of  this  velocity  energy 
could  be  transformed  into  work.  When  designing  a  multi-stage 
turbine,  however,  it  is  necessary  to  find  the  energy  available  for 
each  stage.  This  amount  of  energy  is  affected  by  the  reheating  in 
each  of  the  preceding  stages,  as  well  as  the  drop  in  pressure  in  the 
stage  being  considered. 

For  any  steam  prime  mover,  the  available  energy  is  a  small 
part  of  the  heat  supplied  to  it,  since  a  very  large  part  of  this  heat 
must  necessarily  be  given  up  to  the  exhaust. 


PREPARATION  AND  USE  OF  THE  STEAM  CHARTS  AND 
TABLE  OF  VELOCITIES 

The  general  idea  of  the  main  diagram,  which  consists  of  Plates 
1  to  6,  inclusive,  may  be  readily  obtained  from  the  index  chart. 
It  is  seen  to  be  divided  into  twelve  equal  parts,  the  top  and  bottom 
halves  of  each  section  being  indicated  by  the  subscripts  a  and  b, 
respectively.  These  two  halves  will  be  found  facing  each  other,  so 
that  wherever  this  chart  is  opened,  a  complete  section  may  be  seen. 

The  same  total  heat  scale  is  used  throughout,  but  the  volume 
scale  for  each  section  is  changed  so  that  the  general  relation  of 
each  family  of  curves  to  one  another  will  remain  about  the  same. 

Having  established  the  scale  of  volumes  and  total  heats,  con- 
stant pressure  lines  were  then  plotted  from  the  values  given  by  the 
steam  tables  of  Marks  and  Davis.  In  the  superheated  region  these 
lines  are  slightly  curved,  but  in  the  wet  region  they  are  straight. 
For  those  pressures  not  given  in  the  steam  tables — all  fractional 
pressures  and  those  given  in  inches  of  mercury — the  volumes  and 
total  heats  were  determined  in  two  ways.  For  the  wet  region, 
special  auxiliary  curves  were  drawn  by  computing  for  certain 
qualities  the  total  heats  and  volumes  for  those  pressures  given  by 
the  steam  tables.  From  such  auxiliary  curves  the  desired  values 
could  then  be  determined,  and  the  curves  plotted  in  their  correct 
relative  positions.  In  the  same  manner  the  corresponding  values 
were  found  for  all  fractional  pressures  above  one  pound  in  the 
superheated  region.  For  those  pressures  less  than  one  pound  the 
volumes  in  the  superheated  field  were  found  from  Linde's  equation 
and  the  total  heats  were  determined  by  adding  cpD  to  the  total 
heat  of  the  dry  saturated  steam.  The  values  of  the  specific  heat 
for  these  low  pressures  were  determined  by  assuming  as  correct 
the  specific  heats  used  by  Marks  and  Davis  for  the  pressures  from 
one  to  four  pounds.  A  curve  of  specific  heats  was  then  constructed 
through  these  points  and  extended  into  the  region  of  lower  pres- 
sures. From  such  a  curve  the  specific  heats  were  read.  This  was 
the  best  method  by  which  the  lines  in  the  superheated  region  of 

14 


STEAM   CHARTS   AND   TABLE    OF   VELOCITIES  15 

Plate  6a  could  be  drawn  so  that  they  would  all  continue  as  smooth 
curves. 

For  the  region  of  very  low  pressures,  such  as  those  given  on 
Plate  6a,  it  was  found  that  constant  temperature  lines  would  almost 
coincide  with  the  lines  of  constant  total  heat  when  in  the  super- 
heated region.  It  was,  therefore,  considered  worth  while  to  put 
on  the  scale  of  approximate  temperatures  as  given  on  the  upper 
right-hand  corner  of  this  sheet.  They  will  be  found  to  be  of  service 
when  it  is  desired  to  determine  the  pressure  corresponding  to  a 
certain  temperature  and  specific  volume.  For  a  large  part  of  this 
superheated  region  this  temperature  scale  agrees  with  the  temper- 
ature as  obtained  from  the  degrees  of  superheat  and  the  temper- 
ature of  vaporization.  It  is  not  intended,  however,  to  be  used  as 
an  accurate  means  of  obtaining  the  temperature,  as  the  error  in 
using  it  may  easily  be  one  degree. 

Entropies. — To  draw  accurately  the  lines  of  constant  entropy 
it  was  necessary  to  construct  a  large  total  heat-entropy  diagram 
from  which  could  be  obtained  the  values  of  total  heat  for  the  various 
pressure  lines  and  any  entropy  line.  In  the  superheated  region 
this  was  done  for  each  entropy  line,  but  for  the  wet  region  only 
every  fifth  line  was  obtained  in  this  manner,  as  the  others  could  be 
put  in  by  dividers. 

Qualities. — The  lines  of  constant  quality  were  obtained  by  com- 
puting the  total  heats  for  such  qualities  as  70,  80,  and  90  per  cent., 
and  after  these  were  plotted,  the  intermediate  ones  were  obtained 
by  dividers. 

The  Heat  of  the  Liquid  Curve. — It  was  desired  to  have  the  heat 
of  the  liquid  for  all  pressures,  if  it  could  be  obtained  without  inter- 
fering with  the  other  lines  of  the  chart.  The  lower  left-hand  portion 
of  each  section  was  the  only  space  available,  and  it  was  found  by 
trial  that  this  curve  would  fit  there  very  conveniently.  Since  it 
is  important  to  be  able  to  read  the  heat  of  the  liquid  just  as  accu- 
rately as  the  total  heat,  these  values  should  naturally  be  plotted 
on  scales  of  equal  magnitude.  This  has  been  done  by  supplying 
the  numbers  in  parentheses,  thus  establishing  the  heat  of  the  liquid 
scale. 


16  STEAM   CHARTS   AND   TABLE    OF   VELOCITIES 

Temperature  of  Vaporization.  —  Since  the  temperature  of  dry 
saturated  steam  is  the  same  as  that  of  wet  steam  having  the  same 
pressure,  it  is  possible  to  construct  a  scale  on  any  line  intersecting 
the  constant  pressure  lines  in  the  wet  region,  so  that  such  a  scale 
will  represent  the  temperatures  of  vaporization.  Inasmuch  as  the 
heat  of  the  liquid  is  often  desired  for  some  definite  temperature,  it 
is  natural  to  try  to  place  these  two  curves  as  close  together  as  pos- 
sible. They  were  therefore  combined  by  merely  graduating  the 
heat  of  the  liquid  curve,  already  drawn,  so  that  it  would  also  give 
the  temperature  of  vaporization  for  each  pressure. 

In  order  to  find  the  temperature  of  the  superheated  steam,  it  is 
merely  necessary  to  add  the  degrees  of  superheat  to  the  temper- 
ature of  vaporization. 

Plate  No.  7  is  a  special  addition  to  the  main  heat-volume  chart 
in  order  to  enable  one  to  work  with  unusually  high  superheats  for 
pressures  ranging  from  10  to  45  pounds  per  square  inch  absolute, 
such  as  are  used  when  reheating  steam  to  superheats  of  500  or  600°, 
as  is  done  in  the  Ferranti  turbine.  It  was  not  thought  worth  while 
to  show  this  plate  on  the  Index  Chart. 

Plate  No.  8  gives  the  external  work  in  B.  t.  u.  done  by  a  pound  of 
steam  during  its  formation  at  constant  pressure  from  water  at 
32°  F.,  until  it  reaches  the  state  under  consideration.  For  Plate  8a 
there  are  two  volume  scales  and  for  8b  four  are  used.  The  small 
drawing  in  the  upper  right-hand  corner  is  intended  to  give  the 
external  work  for  the  low  pressures  when  the  quality  is  relatively 
high,  so  that  the  volumes  become  too  great  for  the  main  part  of  8b. 
For  pressures  below  one  pound  absolute,  there  is  a  region  for  which 
no  values  are  given,  but  it  will  seldom,  if  ever,  happen  that  such 
values  will  be  desired. 

Table  of  Velocities.  —  Any  body  having  a  weight  of  one  pound 
and  a  velocity  of  v  feet  per  second,  will  have  kinetic  energy  due  to 
this  velocity  equal  to 


which  is  equal  to  v2 

64.34  X  777.5  B*  t<  U* 


STEAM   CHARTS   AND   TABLE   OF   VELOCITIES  17 

Calling  this  kinetic  energy  in  B.  t.  u.,  E,  the  equation    becomes 
v  =  V64.34  X  777.5  E 
=  223.7 


Then,  for  a  nozzle  which  transforms  the  available  heat  energy  into 
velocity  without  any  losses,  the  velocity  is 


v  —  223.7  Vavailable  energy,  B.  t.  u. 

Table  IV  was  prepared  by  using  this  equation. 

The  available  energy  in  a  steam  nozzle  is  usually  taken  as  the 
total  heat  at  entrance  to  the  nozzle  minus  the  total  heat  for  the  nozzle 
back  pressure,  and  the  same  entropy  as  at  entrance.  This  neglects 
the  small  difference  between  PI  Vwi  and  P2  Vw2  where  P  repre- 
sents pressure  and  Vw  the  volume  of  the  water.  This  difference  is 
usually  far  too  small  to  be  considered  in  actual  cases.  Thus,  taking 
.017  as  the  average  volume  in  cubic  feet  of  a  pound  of  hot  water, 
this  difference  is  equal  to 

017  X  144 

?78     -  (pi  -  p2)  =  .00315  (P!  -  p2)  B.  t.  u.  per  Ib. 

For  a  drop  of  100  pounds  per  square  inch  in  passing  through  the 
nozzle,  this  difference  would,  therefore,  become  only  0.3  B.  t.  u. 
per  pound. 

Using  the  Charts. — Whenever  any  two  of  the  properties  of  super- 
heated steam  are  given,  that  is  sufficient  to  locate  the  state  point 
on  the  chart  from  which  all  of  the  other  values  may  be  determined 
at  once.  This  is  also  true  for  wet  steam,  except  for  the  case  in  which 
the  two  values  given  are  pressure  and  temperature.  These  two 
alone  are  not  sufficient  to  determine  the  state  point,  since  the  tem- 
perature of  wet  steam  is  independent  of  its  quality.  The  problems 
will  supply  many  illustrations  of  the  use  of  the  charts  and  tables. 


ATMOSPHERIC   PRESSURE   AND   BAROMETRIC 
CORRECTIONS 

The  average  pressure  of  the  atmosphere  at  sea  level  is  about 
14.7  pounds  per  square  inch,  and  for  many  purposes  this  value  is  all 
that  is  needed.  On  the  other  hand,  it  is  often  necessary  to  determine 
carefully  the  atmospheric  pressure  at  the  time  and  place  desired. 
This  is  usually  done  by  measuring  the  height  of  a  column  of  mercury 
which  is  just  balanced  by  the  atmosphere.  Pressures  less  than 
atmospheric  are  also  often  determined  by  means  of  mercury  col- 
umns. All  such  measurements  may  be  accurately  made,  if  care  is 
exercised  in  obtaining  the  readings  and  in  applying  the  proper  correc- 
tions. 

The  Standard  Atmospheric  Pressure  *  is  equivalent  to  the 
height  of  a  column  of  mercury  760  millimetres  high,  at  a  temper- 
ture  of  0°  C.,  at  sea  level  and  45°  latitude.  Reducing  this  to  the 
inch  basis  the  standard  atmospheric  pressure  becomes 

760  X  .03937  =  29.9212  inches 

at  32°,  at  sea  level  .and  latitude  of  45°.  Since  the  engineer  is  not 
concerned  with  any  readings  of  pressure  closer  than  the  thousandth 
of  an  inch  of  mercury,  and  very  often  only  to  the  nearest  hundredth, 
this  is  equivalent  to  the  value  usually  given,  viz.,  29.921  inches  at 
32°  F. 

The  Thirty-Inch  Barometer. — When  measuring  the  pressure  in 
a  condenser,  engineers  often  speak  of  the  vacuum  referred  to  a 
thirty-inch  barometer.  The  meaning  of  this  expression  may  not 
always  be  the  same,  as  it  may  be  interpreted  differently.  However, 
a  logical  meaning  and  one  quite  generally  used  f  is  that  30  inches 

*  The  exact  value  adopted  for  international  use  by  the  Third  General  Conference  on  Weights  and 
Measures  is  that  of  a  column  of  mercury  760  millimetres  high,  the  mercury  being  at  a  temperature  of 
0°  C.  and  the  acceleration  of  gravity  being  980.665  centimetres  per  second  per  second.  See  Vol.  XII, 
p.  66,  of  the  "Travaux  et  Memoirs  du  Bureau  International  des  Poids  et  Mesures." 

t  See  Genhardt's  "Steam  Power  Plant  Engineering,"  p.  463,  4th  edition.  Also  "Steam  Tables  for 
Condenser  Work  "  by  the  Wheeler  Condenser  and  Engineering  Company,  page  5. 

18 


ATMOSPHERIC   PRESSURE   AND   BAROMETRIC   CORRECTIONS        19 

of  mercury  at  some  definite  temperature  will  mean  the  same  pressure 
as  do  the  29.921  inches  at  32°  F.  To  find  this  temperature  the 
equation 

30.00  =  29.921  [l  +  a  (t  -  32)] 

may  be  used  where 
t  =  the  required  temperature,  and 

a  =  .000101  =  coefficient  of  cubical  expansion  of  mercury  per 
degree  F. 

From  this  equation 

t  =  58.1°  F. 

Had  the  standard  pressure  been  taken  as  29.92  inches  instead  of 
29.921  inches,  this  temperature  would,  from  the  above  equation, 
become  58.4°  F. 

In  making  the  above  correction  for  temperature,  it  was  assumed 
that  the  scale  by  which  the  mercury  is  measured  is  correct  at  this 
temperature.  This  is  often  the  case,  as  when  very  short  scales  or 
scales  made  of  non-expansive  materials  are  used.  However,  full- 
length  brass  scales,  or  their  equivalent,  are  sometimes  used.  If 
such  a  brass  scale  be  correct  at  62°  F.,  as  is  usually  intended,  the 
temperature  at  which  30  such  inches  represent  the  standard  pressure 
may  be  found  from  the  equation: 


where       a  =  .000101  as  above, 

and          b  =  .00001  =  coefficient  of  expansion  of  brass. 
The  solution  of  this  equation  will  give,    t  =  57.6°  F. 
The  difference  in  the  above  three  values  for  this  temperature  is 
not  of  great  importance.    In  all  the  steam  charts  used  in  this  book 
the  temperature  of  the  mercury  has  been  taken  as  58.1°  F.,  as  that 
would  seem  to  serve  a  more  general  use  than  does  the  value  57.6° 
derived  on  the  assumption  of  a  full-length  brass  scale  having  a 
definite  coefficient  of  expansion. 

Correction  Due  to  Temperature.  —  In  order  to  facilitate  the 
reduction  of  readings  of  a  mercury  column  at  any  temperature  to 
the  temperature  of  58.1°,  Plate  9a  has  been  prepared.  It  is  correct 


20        ATMOSPHERIC    PRESSURE    AND   BAROMETRIC    CORRECTIONS 

for  those  cases  in  which  a  non-expansive  scale  is  used,  and  may  also, 
of  course,  be  used  indirectly  to  correct  to  any  other  temperature 
desired,  say  32°. 

When  a  Full-Length  Brass  Scale  is  used  to  measure  the  height 
of  a  column  of  mercury  of  some  considerable  magnitude,  such  as 
a  barometer,  and  the  temperature  of  this  scale  is  not  close  to  62°, 
the  correction  should  be  made  to  the  scale  as  well  as  to  the  mercury. 

Thus,  for  barometers  having  such  scales,  it  may  be  desired  to 
reduce  the  reading  to  32°  for  the  mercury  and  62°  for  the  brass 
scale.  The  following  values  *  give  this  correction  for  a  column  30 
inches  high.  Hence,  for  any  other  column,  say  h,  it  is  only  necessary 

to  multiply  by  (— ) 


Temp.0  F. 

CORRECTION  IN  INCHES 
(For  30  inches) 

Temp.0  F. 

CORRECTION  IN  INCHES 
(For  30  inches) 

100 

Subtract  .193 

40                       Subtract  .031 

95 
90 

.180 
.166 

35 
32 

.017 
.009 

85 

.153 

30 

.004 

80 
75 

.139 
.126 

28.5 
25 

Add         .000 
.010 

70 

.112 

20 

.024 

65 
62 

.099 
.091 

15 

10 

.037 
.051 

60 

.085 

5 

.065 

58.1 

55 

.080 
.072 

0 
-  5. 

.078 
.092 

50 

.058 

-10 

.106 

45 

.045 

-15 

.119 

Variation  of  Atmospheric  Pressure  with  Altitude. — It  may  hap- 
pen that  it  is  not  convenient  or  practicable  to  have  a  barometer  at 
the  same  place  at  which  it  is  desired  to  measure  the  pressure  of  the 
atmosphere.  It  does  not  require  much  of  a  difference  in  elevation  in 


*From  Table  1,  "Circular  F,  U.  S.  Weather  Bureau  No.  472.' 


ATMOSPHERIC    PRESSURE    AND   BAROMETRIC    CORRECTIONS        21 

order  to  make  quite  a  variation  in  atmospheric  pressure.  To  obtain 
the  true  pressure  at  some  elevation  which  is  not  extremely  different 
from  that  at  which  a  barometer  is  read,  it  is  only  necessary  to 
observe  the  temperature  of  the  atmosphere  and  obtain  the  alti- 
tude of  each  place.  Then,  by  means  of  Plate  9b,  this  correction 
may  be  readily  obtained.  This  plate  *  was  intended  to  serve  only 
in  those  cases  in  which  the  change  in  altitude  might  be  100  feet  or 
less.  But  after  completion  it  was  found  to  agree  with  the  correc- 
tions obtained  from  the  above  table  for  variations  in  altitude  of  as 
much  as  1,000  feet.  The  average  altitude  is  to  be  used  when  ob- 
taining the  correction  from  this  plate. 

Reduction  to  Standard  Gravity. — When  it  is  desired  to  make  a 
true  comparison  of  pressures  determined  by  heights  of  mercury 
columns,  it  is  necessary  to  reduce  them  to  a  common  standard  of 
gravity.  The  standard  usually  adopted  for  this  is  that  at  sea  level 
and  a  latitude  of  45°.  Table  If  will  give  these  corrections  for  the 
various  latitudes. 

The  variation  of  gravity  due  to  change  in  altitude  is  usually  much 
too  small  to  require  correction  of  mercury  columns.  This  correc- 
tion is  proportional  to  the  height  of  the  mercury  and  for  a  30-inch 
column  amounts  to  about  .0018-inch  for  each  1,000  feet  above 
sea-level.  It  is,  of  course,  to  be  subtracted.  It  is  important  not 
to  confuse  this  correction  with  the  variation  of  the  pressure  of 
the  atmosphere  itself,  due  to  change  in  altitude,  as  has  already 
been  given. 

Correction  Due  to  Capillarity. — The  capillary  action  between 
mercury  and  glass  will  cause  a  depression  of  the  column  if  the 
pressure  is  constant  so  that  the  mercury  becomes  stationary.  The 
correction  to  be  made  for  this  action  may  be  very  large  in  case 
the  tube  or  the  mercury  is  dirty  or  if  a  very  small  tube  is  used. 
The  height  of  the  meniscus  and  the  diameter  of  the  tube  are  the 
two  most  important  factors  required  to  determine  the  amount  of 
this  correction.  Table  II  will  give  these  corrections  for  nearly 
all  cases. 

*  Prepared  from  Table  21,  Vol.  II,  "Report  of  the  Chief  of  the  Weather  Bureau,  1900-1901." 
t  Abridged  from  Table  101  of  the  "Smithsonian  Tables." 


22        ATMOSPHERIC  PRESSURE  AND  BAROMETRIC   CORRECTIONS 

Reduction  of  Mercury  Column  to  Pounds  per  Square  Inch. — For 

the  most  important  work  in  connection  with  the  measurement  of 
steam  pressures  by  means  of  mercury  columns,  it  is  not  usually 
necessary  to  convert  the  reading  to  any  other  unit.  Thus  the 
absolute  pressure  hi  a  condenser  being  measured  by  the  difference 
in  the  heights  of  two  mercury  columns,  it  is  convenient  and  proper 
to  use  the  inch  of  mercury  as  the  unit  to  express  such  pressures. 

Density  of  Mercury. — When  it  is  necessary  to  convert  inches  of 
mercury  to  pounds  per  square  inch,  Table  III  may  be  used.  This 
table  was  prepared  by  using  the  following  equation: 

One  cubic  inch  at  32°  F.  is  equal  to 

1  +  (.000101)  (t  -  32)  cu.  in.  at  t°  F. 

The  density,  0.49117  pounds  per  cubic  inch  at  32°  F.,  was  obtained 
from  the  equivalent,  13.59545  grams  per  c.c.  at  0°  C.*  Then  these 
values  for  all  temperatures  were  afterward  compared  and  found  to 
agree  with  Table  76  of  the  Smithsonian  Tables. 

*  This  is  the  value  given  by  Table  19  of  "Landolt  and  Bernstein  Physikalisch-Chemische  Tabellen." 


44-. 


fc 


k" 


& 


$* 

i     "h 
^^ 


p4 
^ 


1    1 


ol 

Of 


K 


^  ii^ 


41 


-K 


fci 


4* 


1 


(23 


Specific-  Volume 


(24)  PLATE  IA 

Specific  Volume,  Cu.  Ft.  per  Lb. 

1.2  1.4  1.6  1.8  2.0  2.2  2.4  2.6  2.8  3.0  3.2  3.4  3.6  3.8 


1510 


1490 


1.2  1.4  1.6  1.8  2.0  2.2  2.4  2.6  2.8  3.0  3.2  3.4  3.6  3.8 


Specific  Volume,  Cu.  Ft.  pet  Lb. 


1310 


1300 


PLATE  IB 


1.2  1.4  1.6 


(25) 


Specific  Volume,  Cu.  Ft  per  Lb . 

2.0  2.2  2.4  2.6  2.8  3.0  3.2  3.4  3.6  3.8 


1.6  1.8  2.0  2.2  2.4  2.6  2.8  3.0  3.2  3.4  3.6  3.8 

Specific  Volume,  Cu.  Ft.  per  Lb. 


(26) 


Specific  Volume,  Cu.  Ft.  perLb. 

7  8 


PLATE  2A 


1240 


6789 

Specific  Volume,  Cu.  Ft.  per  Lb. 


10 


PLATE  2s 


(27) 


Specific  Volume,  Cu.  ft.  per  lb. 


10 


1230 


1220 


1210 


1200 


1030 
U30) 


7  8  9 

Specific  Volume,  Cu.  ft.  per  lb. 


10 


(28) 


PLATE  SA 


Specific  Volume,  Cu.  Ft.  per  Lb. 

12     14      16     18     20     22     24     26      28     30 


32      34     36 


1350 


U340 


1330 


1320 


1310 


1300 


1290 


1280 


1270 


1260 


1250 


1240 


1230 


1220 


1210 


1200 


1190 


1180 


1170 


1160 


1150 


1350 


1340 


1170 


1160 


1150 


14     16     18 


20     22     24     26     28     30 

Specific  Volume,  Cu.Ft.  perLb. 


32  34  36  38 


PLATE  3s 


(29) 


1140 


1130 


1120 


1140 


1130 


Specific  Volume,  Cu.  Ft.  per  Lb 


12     14     16     18     20     22     24     26      28     30     32     34     36 


12     14     16     18     20     22     24     26     28     30     32     34     36     38 


Specific  Volume,  Cu.Ft.  perLb. 


940 
(159) 


940 


(30) 


40 


50 


Specific  Volume,  Cu.  Ft.  per  Lb. 

70         80         90        100 


PLATE  4A 

120       130 


1280 


1270 


1260 


1250 


1280 


1270 


1260 


1080 


50 


70  80  90  100 

Specific  Volume,  Cu.  Ft.  per  Lb. 


120 


130 


PLATE  4s 


50 


60 


Specific  Volume,  Cu.  ft.  per  Ib. 

70        80         90        100 


120 


(31) 


130 


1070 


1060 


1050 


1040 


50 


70        80        90 

Specific  Volume,  Cu.  ft.  per  Ib. 


100 


110 


120 


130 


(32) 


150 


200 


Specific  Volume,  Cu.  Ft.  per  Lb. 

250          300  350 


PLATE 

450 


1210 


1200 


1190 


1010 


1010 


200 


250  300  350 

Specific  Volume,  Cu.  Ft.  per  Lb. 


400 


450 


PLATE  SB 

150 


(33) 


Specific  Volume,  Cu.  Ft.  per  Lb. 

250  300  350 


400 


450 


1000 


990 


980 


200 


250  300  350 

Specific  Volume,  Cu.  Ft.  per  Lb.> 


400 


450 


(34) 

500          600 


PLATE  GA 


Specific  Volume,  Cu.  Ft.  per  Lb. 

700     800    900    1000    1100    1200    1300    1400    1600    1800 


2000    2200 


1140  H 


950 


940 


500     600     700    800     900    1000    1100    1200    1300    1400    1600    1800    2000 

Specific  Volume,  Cu.  Ft.  per  Lb. 


2200 


PLATE  6s 

500  600  700 


(35) 


Specific  Volume,  Cu.  Ft.  perLb. 

800  900  1000          1100         1200          1300         1400          1600         1800        2000          2200 


930 


920 


910 


810 


800 


780 


770 


760 


750 


740 


730 


700          800          900          1000         1100         1200         1300         1400         1600         1800         2000         2200 
Specific  Volume,  Cu.  Ft.  per  Lb. 


(36) 


PLATE  7 


16  18 


22 


Specif ic  Volume,  Cu.  Ft.  perLb. 

24     26     28     30  40 


50 


1440 


1430 


1440 


1430 


1420 


1410 


1400 


1390  r 


1380  ma 


1370  5 


1360- 


1350 


1340 


1330 


1320 


1310 


1300 


1290 


1300 


=  1290 


16     18     20     22 


24     26     28     30  40 

Specific  Volume,  Cu.  Ft.  per  Lb. 


60 


THIS  PLATE  IS  TO  SUPPLEMENT  PLATES  3A  AND  4A  FOR  THOSE 
EXCEPTIONAL  CASES  IN  WHICH  EXTREMELY  HIGH  SUPERHEAT 
IS  USED  FOR  COMPARATIVELY  LOW  PRESSURES. 


THE  TWO  FOLLOWING  PLATES  GIVE  THE  EXTERNAL  WORK  DONE 
DURING  THE  CONSTANT  PRESSURE  FORMATION  OF  STEAM  FROM 
WATER  AT  32°  F. 


(37) 


(38) 


PLATE  SA 


Specific  Volume,  Cu.  Ft.  per  Lb. 

4568  10 


12 


14 


16 


18 


160 


150 


140 


10 


45  6  8  10  12 

Specific  Volume,  Cu.  Ft.  per  Lb. 


14 


16 


18 


PLATE  8s 


(39) 


160 


300  400  600  800 

Specific  Volume,  Cu.  Ft.  per  Lb. 


20 


10 


60  70  80  100  120 

Specific  Volume,  Cu.  Ft.  per  Lb. 


140 


160 


(40)  PLATE  QA 

Showing  correction  of  Mercury  Column  due  to  Temperature,  when  scale 
is  correct  at  the  observed  temperature 


Observed  reading  in  Inches  of  Mercury 

10  15  20 


25 


10  15  20 

Observed  reading  in  Inches  of  Mercury 


30 


PLATE  9s 
Showing  correction  of  Barometric  Reading  due  to  change  in  Elevation 


1000 


2000 


Altitude  in  Feet 

3000  4000 


3000  4000 

Altitude  in  Feet 


(41) 


TABLE  I 

SHOWING    CORRECTIONS    TO    REDUCE    BAROMETRIC    READINGS    TO    45°    LATITUDE* 


For  These  Lati- 
tudes the  Cor- 
rection is  to  be 
Subtracted 

Barometer  Reading  in  Inches  of  Mercury 

Frfr  These 
Latitudes  the 
Correction  is 
to  be  Added 

22 

23 

24 

25 

26 

27 

28 

29 

30 

0° 

.059 

.061 

.064 

.067 

.069 

.072 

.074 

.077 

.080 

90° 

.      10° 

.055 

.058 

.060 

.063 

.065 

.068 

.070 

.073 

.075 

80° 

20° 

.045 

.047 

.049 

.051 

.053 

.055 

.057 

.059 

.061 

70° 

30° 

.029 

.031 

.032 

.033 

.035 

.036 

.037 

.039 

.040 

60° 

40° 

.010 

.011 

.011 

.012 

.012 

.012 

.013 

.013 

.014 

50° 

45° 

.000 

.000 

.000 

.000 

.000 

.000 

.000 

.000 

.000 

45° 

*  Abridged  and  rearranged  from  Table  101  of  the  Smithsonian  Tables. 

TABLE  II 

CORRECTION   OF   THE  BAROMETER   FOR  CAPILLARITY 

Correction  to  be  added  in  inches 


Diameter 
of 

] 

ieight  of  1 

Meniscus  i 

ti  Inches 

Tube 
in  Inches 

.01 

.02 

.03 

.04 

.05 

.06 

.07 

.08 

.15 

.024 

.047 

.069 

.092 

.116 

.20 

.011 

.022 

.033 

.045 

.059 

.079 

.25 

.006 

.012 

.019 

.028 

.037 

.047 

.059 

.... 

.30 

.004 

.008 

.013 

.018 

.023 

.029 

.035 

.042 

.35 

.005 

.008 

.012 

.015 

.019 

.022 

.027 

.40 

.004 

.006 

.008 

.010 

.012 

.014 

.016 

.45 

.... 

.... 

.003 

.005 

.007 

.008 

.010 

.012 

.50 

.... 

.... 

.002 

.004 

.005 

.006 

.006 

.007 

.55 

.001 

.002 

.003 

.004 

.005 

.005 

t  From  Table  103  Smithsonian  Tables,  modified  by  giving  the  correction  only  to  the  nearest  thousandth 
of  an  inch. 


TABLE  III 

DENSITY   OF  MERCURY 


Temperature 

Pounds  per 
Cubic  Inch 

Temperature 

Pounds  per 
Cubic  Inch 

0 

.4928 

58.1 

.4899 

10 

.4923 

60 

.4898 

20 

.4918 

70 

.4893 

30 

.4913 

80 

.4888 

32 

.4912 

90 

.4883 

40 

.4907 

100 

.4878 

50 

.4903 

110 

.4873 

J  See  page  22. 


(42) 

TABLE  IV 

SHOWING  THE  THEORETICAL  VELOCITIES  ATTAINED 

BY  STEAM  EXPANDING  ADIABATICALLY  IN  A 

FRICTIONLESS  NOZZLE 

The  velocities  are  given  in  feet  per  second  for  each  B.  t.  u.  up  to  599. 
012345678 


0 

0 

224 

316 

387 

447 

500 

548 

592 

633 

671 

10 

707 

742 

775 

806 

837 

866 

895 

922 

949 

975 

20 

1001 

1026 

1050 

1073 

1097 

1120 

1141 

1163 

1184 

1205 

30 

1226 

1246 

1266 

1285 

1304 

1323 

1342 

1361 

1379 

1397 

40 

1415 

1433 

1450 

1467 

1484 

1501 

1517 

1533 

1550 

1566 

50  1582  1598  1613  1628  1643  1658  1673  1688  1703  1718 

60  1732  1747  1761  1775  1789  1803  1817  1831  1844  1858 

70  1872  1885  1898  1911  1924  1937  1950  1963  1976  1988 

80  2000  2013  2026  2038  2050  2062  2074  2086  2098  2110 

90  2122  2134  2146  2158  2169  2180  2191  2202  2214  2226 

100  2237  2248  2259  2270  2281  2292  2303  2314  2325  2336 

110  2346  2356  2367  2378  2389  2399  2409  2419  2430  2440 

120  2450  2460  2470  2480  2490  2500  2511  2521  2531  2540 

130  2550  2560  2570  2580  2590  2600  2609  2619  2628  2637 

140  2647  2657  2666  2675  2684  2694  2703  2712  2721  2730 

150  2740  2749  2758  2767  2776  2785  2794  2803  2812  2821 

160  2830  2839  2848  2857  2866  2874  2882  2891  2900  2908 

170  2917  2925  2934  2942  2951  2960  2968  2976  2984  2993 

180  3001  3010  3018  3026  3034  3042  3050  3059  3067  3075 

190  3083  3091  3100  3108  3116  3124  3132  3140  3148  3156 

200  3164  3172  3180  3188  3196  3204  3211  3219  3227  3234 

210  3241  3249  3257  3265  3273  3280  3288  3296  3303  3310 

220  3318  3325  3332  3340  3348  3355  3363  3370  3377  3384 

230  3392  3400  3407  3414  3422  3430  3437  3444  3451  3458 

240  3465  3473  3480  3487  3494  3501  3508  3516  3523  3530 

250  3537  3544  3551  3558  3565  3572  3579  3586  3593  3600 

260  3607  3614  3620  3627  3634  3641  3648  3655  3662  3669 

270  3676  3683  3689  3696  3703  3710  3717  3723  3730  3737 

280  3743  3750  3757  3763  3770  3777  3783  3790  3796  3803 

290  3810  3817  3823  3829  3835  3842  3849  3855  3861  3868 


(43) 

TABLE  IV 

SHOWING  THE  THEORETICAL  VELOCITIES  ATTAINED 

BY  STEAM  EXPANDING  ADIABATICALLY  IN  A 
FRICTIONLESS   NOZZLE 

The  velocities  are  given  in  feet  per  second  for  each  B.  t.  u.  up  to  599. 

°  12345678 

300    3874  3881   3888   3894   3900   3907   3913  3920  3926 

310    3939  3946   3952   3958   3964   3970   3976  3982  3989 

320    4002  4008   4014   4020   4027   4033   4039  4045  4051 

330    4063  4070   4076   4082   4088   4094   4100  4107  4113 

340    4125  4131   4137   4143   4149   4155   4161  4167  4173  41' 

350    4185  4191   4197   4203   4209   4215   4221  4227  4233  42^ 

360    4245  4251   4257   4263   4268   4274   4280  4286  4291  42! 

370    4302  4308   4314   4320   4326   4332   4338  4344  4350  43^ 

380    4361  4367   4372   4378   4383   4389   4395  4401  4407  44: 

390    4418  4424   4430   4435   4440   4446   4451  4457  4462  44< 

400    4473  4479   4485   4490   4496   4502   4508  4513  4519  45< 

410    4530  4536   4541   4547   4552   4558   4563  4569  4574  45! 

420    4585  4590   4596   4601   4607   4612   4617  4623  4628  46! 

430    4639  4644   4650   4655   4661   4666   4671  4677  4682  46! 

440    4693  4698   4703   4709   4714   4719   4724  4729  4735  47< 

450    4745  4750   4755   4761   4766   4771   4776  4781  4787  47! 

460    4797  4802   4808   4813   4818   4823   4828  4833  4839  48' 

470    4849  4854   4859   4865   4871   4875   4880  4885  4891  48! 

480    4901  4906   4911   4917   4922   4927   4932  4937  4942  49^ 

490    4952  4957   4962   4967   4972   4977   4982  4987  4992  49! 

500    5002  5007   5012   5017   5022   5027   5032  5037  5042  50^ 

510    5052  5057   5062   5067   5072   5077   5082  5087  5091  50< 

520    5101  5106   5111   5116   5121   5126   5131  5136  5140  51^ 

530    5150  5155   5160   5164   5169   5174   5179  5184  5188  51! 

540    5198  5203   5208   5212   5217   5222   5227  5232  5236  52^ 

550    5246  5251   5256   5260   5265   5270   5275  5280  5284  52* 

560    5294  5299   5303   5308   5312   5317   5322  5327  5331  53^ 

570    5341  5346   5350   5355   5359   5364   5369  5373  5378  53* 

580    5387  5392   5397   5401   5406   5411   5416  5420  5425  54S 

590    5434  5439   5443   5448   5452   5457   5461  5466  5470  54^ 


PROBLEMS 

1.  Find  the  total  heat,  volume,  entropy,  and  temperature  of  a 
pound  of  steam,  having  an  absolute  pressure  of  75  pounds  per 
square  inch,  and  superheated  280°  F. 

Solution:  Referring  to  the  index  chart,  it  will  be  seen  that  for 
this  pressure  Plates  2a  and  2b  should  be  used. 
From  Plate  2a,  we  read  directly: 

Total  heat  =  1320.7  B.  t.  u. 
Volume       =  8.28  cu.  ft. 
Entropy      =  1.78 

From  Plate  2b,  we  read  from  the  temperature  of  vaporization 
curve  near  the  bottom  of  the  Plate: 

Temp,  of  vaporization  for  75  Ibs.  per  sq.  in.  abs.  =  307.6°  F. 
Hence,  for  280°  superheat, 

Temperature  =  280  +  307.6  =  587.6°  F. 

2.  Find  the  total  heat,  volume,  entropy,  temperature,  and  heat 
of  the  liquid  of  a  pound  of  steam  having  an  absolute  pressure    of 
180  pounds  per  square  inch  and  a  quality  of  98%. 

Solution:  From  Plate  Ib,  we  read  directly: 

Total  heat  =  1179.3  B.  t.  u. 
Volume  =  2.48  cu.  ft. 
Entropy  =  1.534 
Temperature  =  373.1°  F. 
Heat  of  the  liquid  =  345.5  B.  t.  u. 

3.  Eight  pounds  of  steam  are  confined  in  a  space  of  800  cubic 
feet.      If  the  quality  of  this  steam  is  92%,  find  its  pressure  and 
temperature. 

45 


46  PROBLEMS 

Solution: 

Specific  volume  =  -g-  =  100  cu.  ft.  per  Ib. 

Turning  to  Plate  4b  and  following  up  this  specific-volume  line, 
100  cu.  ft.,  until  we  intersect  the  quality  of  92%,  we  read: 

Pressure  =  3.3  pounds  per  sq.  in.  absolute 

By  running  down  this  pressure  line,  which  is  approximately  half- 
way between  the  pressure  lines  marked  3.2  and  3.4  pounds  per 
square  inch,  until  we  intersect  the  curve  graduated  to  give  temper- 
atures of  vaporization,  we  read: 

Temperature  =  145.5°  F. 

4.  The  barometer  reads  30.05  inches  of  mercury  at  a  tempera- 
ture of  91°  F.  The  mercury  column  attached  to  the  condenser  reads 
28.6  inches  at  a  temperature  of  110°  F.  Find  the  absolute  pressure 
in  inches  of  mercury  in  the  condenser  at  the  standard  temperature 
of  58.1°  F.* 

Solution:  By  referring  to  Plate  9a,  we  read: 


\t  r 


For  30.05"  at  91°  correction  is  -  .10' 
Hence  barometer  reading  at  58.1°  F.  =  30.05  -  .10  =  29.95" 
Likewise  the  correction  for  the  vacuum  reading  of  28.6"  at  110° 
F.  is  seen  to  be  -  0.15". 
Therefore  vacuum  reading  at  58.1°  =  28.6  -  .15  =  28.45"  and 

absolute  pressure  =  29.95 -28.45  =  1.5"  Hg.  at  58.1°. 

5.  One  pound  of  steam  expands  at  constant  entropy  of  1.69 
until  condenser  pressure  is  attained.  The  manometer  on  this 
condenser  reads  29.50  inches  of  mercury  at  a  temperature  of  92°  F., 
and  the  barometer  reading  is  30.04  inches  of  mercury  at  38°  F.  Find 
the  total  heat,  volume,  quality,  and  temperature  of  this  steam. 

*  This  is  the  temperature  at  which  30  inches  of  mercury  are  equal  to  a  standard  atmosphere,  which  is 
defined  as  being  equal  to  29.921  inches  of  mercury  at  32°  F.  See  page  18  for  further  discussion  of  this 
subject. 


PROBLEMS  47 

Solution:  It  is  first  necessary  to  obtain  the  absolute  condenser 
pressure,  measured  in  inches  of  mercury  at  the  temperature  58.1°, 
which  is  the  temperature  for  which  the  mercury  readings  are  given 
on  the  steam  chart. 

From  Plate  9a,  we  read: 

Barometer,  30.04"  at  38°  =  30.04  +  .06   or   30.10"   at 

58.1°  F. 
Vacuum,  29.50"  at  92°  =  29.50  -  .10  or  29.40"  at 

58.1°  F. 
Therefore  absolute  pressure  in-  condenser  must  be 

30.10  -  29.40  =  0.7  inches  of  mercury  at  58.1°  F. 

Now,  referring  to  Plate  6b,  we  read  for  an  entropy  of  1.69  and  a 
pressure  of  0.7  inches  of  mercury,  the  following  values: 

Total  heat  =  891  B.  t.  u. 

Volume  =  745  cu.  ft. 

Quality  =  81.1% 

Temperature  =  68.6°  F. 

6.  How  many  pounds  of  steam  are  contained  in  a  steam  pipe 
which  is  12  inches  in  diameter  and  200  feet  long,  if  the  average  pres- 
sure in  this  pipe  is  100  pounds  per  square  inch  absolute,  and  the 
average  temperature  is  433.5°  F.? 

Solution:  By  referring  to  Plate  Ib,  we  find  that  the  tempera- 
ture of  vaporization  of  steam  having  an  absolute  pressure  of  150 
pounds  per  square  inch  is  358.5°  F. 

Hence  degrees  of  superheat  =  433.5  -  358.5  =  75°  F.  Then 
from  the  same  plate,  for  a  pressure  of  150  pounds  per  square  inch 
and  75°  superheat,  we  read: 

Specific  volume  =  3.4  cu.  ft.  per  pound. 
Hence  weight  of  steam  in  this  pipe  must  be 

volume  of  pipe  1  X  .7854  X  200 

specific  volume  of  steam  =  ~3A~  t6'2  pounds' 

7.  Find  the  number  of  B.  t.  u.  which  are  required  to  do  the 
external  work  during  the  formation  of  a  pound  of  steam  at  a  con- 


48  PROBLEMS 

stant  pressure-  of  60  pounds  per  square  inch  absolute  from  water 
at  32°  F.  until  it  is  superheated  210°  F. 

Solution:  Turning  to  Plate  8a,  at  the  intersection  of  the  60- 
pound  line  and  the  210°  superheat  line,  we  read: 

External  work  done  during  this  constant  pressure  change 
=  105  B.  t.  u. 

8.  Find  the  intrinsic  heat  of  the  steam  in  the  final  state  given 
in  the  previous  problem. 

Solution:  For  an  absolute  pressure  of  60  pounds  per  square 
inch  and  210°  superheat,  from  Plate  2a  we  read: 

Total  heat  =  1281  B.  t.  u. 

Then,  since  the  intrinsic  heat  is  merely  that  heat  contained 
within  the  steam  itself,  we  may  obtain  it  easily  by  subtracting 
from  the  total  heat  the  external  work  done  by  the  steam  during  its 
constant  pressure  formation  from  water  at  32°  F.  This  external 
work,  from  previous  problems,  is  105  B.  t.  u. 

Hence,  for  this  case, 

Intrinsic  heat  =  1281  -  105  =  1176  B.  t.  u. 

9.  If  a  pound  of  dry  saturated  steam  is  heated  at  a  constant 
pressure  of  200  pounds  per  square  inch  absolute,  until  its  total  heat 
is  increased  by  10%,  find  how  many  degrees  it  has  been  superheated 
and  the  per  cent,  increase  in  its  volume. 

Solution:  From  Plate  Ib,  we  find: 

Initial  total  heat  =  1198.2  B.  t.  u. 

Initial  volume  =  2.29  cu.  ft. 
Hence  final  total  heat  =  1198.2xl.10  =  1318  B.  t.  u.    From 

Plate  la,  for  this  total  heat  and  a  pressure  of  200,  we  read: 
Final  volume  =  3.11  cu.  ft. 
Final  superheat  =  221°. 

3  11  —  2  29 
Therefore  per  cent,  increase  in  volume  =  -     ^  29     "  =  35.8%. 


PROBLEMS  49 

10.  Steam  having  an  absolute  pressure  of  80  pounds  per  square 
inch  and  a  quality  of  98%  is  passed  through  a  superheater  on  its 
way  to  an  engine.  If  the  steam  leaving  the  superheater  has  a  tem- 
perature of  450°  F.,  find: 

(a)  Degrees  of  superheat, 

(b)  Heat  added  per  pound  by  superheater, 

(c)  Specific  volume  of  the  superheated  steam, 

(d)  What  per  cent,  of  the  heat  added  by  this  superheater  was 
required  to  complete  the  vaporization. 

Solution:  For  steam  having  an  absolute  pressure  of  80  pounds 
per  square  inch,  we  find,  from  Plate  2b : 

Temperature  of  vaporization  =  312°  F. 
Hence 

(a)  Degrees  of  superheat  =  450  -  312  =  138°  F. 

Also  from  this  same  plate,  for  the  given  pressure  and  quality, 
we  observe: 

Initial  total  heat  =  1164  B.  t.  u. 

Then,  by  running  along  the  80-pound  pressure  line  until  it  inter- 
sects the  138°  superheat  line,  we  find,  from  Plate  2a: 

Final  total  heat  =  1253  B.  t.  u. 
Hence 

(b)  Heat  added  by  superheater  =  1253  -  1164  =  89  B.  t.  u. 
Also  from  Plate  2a,  we  may  read : 

(c)  Specific  volume  =  6.65  cu.  ft.  per  pound. 

For  the  given  pressure  we  may  read,  from  Plate  2b: 

(d)  Total  heat  of  dry  saturated  steam  =  1182.3  B.  t.  u. 
Hence  the  heat  required  to  complete  the  vaporization  of  the 

wet  steam  is 

1182.3  -  1164  =  18.3  B.  t.  u. 

Therefore  the  portion  of  the  heat  which  was  added  by  the  super- 
heater in  order  to  complete  the  vaporization  is 

^=20.6%          ..:.;".    • 


50  PROBLEMS 

v  1L  If  one  pound  of  steam  expands  adiabatically  from  an  ab- 
solute pressure  of  100  pounds  per  square  inch  and  145°  superheat, 
to  an  absolute  pressure  of  1.4  inches  of  mercury,  find: 

(a)  Final  quality. 

(b)  Final  volume. 

(c)  Ratio  of  expansion. 

(d)  Work  done. 

Solution:  From  Plate  2a,  for  the  100  pound  line  and  145° 
superheat,  we  read: 

Initial  total  heat  =  1262.3  B.  t.  u. 
Initial  entropy      =  1 . 69 
Initial  volume      =  5 . 45  cu.  ft. 

Then  turning  to  Plate  5b,  and  running  along  the  entropy  line 
1.69  until  it  intersects  the  pressure  line  1.4  inches  of  mercury,  we 
read: 

Final  total  heat    =  925  B.  t.  u. 

(a)  Final  quality        =83.4% 

(b)  Final  volume        =  397  cu.  ft. 
From  these  results  we  have: 

/  x  -D  x-      *  •  Final  volume        397       nc. 

(c)  Ratio  of  expansion  ==  Initial  volume  ==  5^5  «  72.8 

In  order  to  obtain  the  work  done  during  this  expansion,  it  is 
necessary  to  obtain  the  intrinsic  heats  for  the  initial  and  final 
states,  since  the  work  done  during  adiabatic  expansion  is  equal 
to  the  loss  of  intrinsic  heat.  The  intrinsic  heat  for  any  state  may 
be  obtained  by  subtracting  the  external  work,  as  obtained  from 
Plate  8,  from  the  total  heat. 

For  the  initial  pressure  of  100  pounds  and  superheat  of  145°, 
from  Plate  8a,  we  read: 

The  constant  pressure  external  work  =  100.3  B.  t.  u. 
Hence 

Initial  intrinsic  heat  =  1262.3-100.3  =  1162  B.  t.  u. 


PROBLEMS 


51 


For  the  final  pressure  of  1.4"  Hg.  and  a  quality  of  83.4%,  we 
read  from  the  upper  right-hand  corner  of  Plate  8b : 

The  constant  pressure  external  work  =  50  B.  t.  u. 
Hence 

Final  intrinsic  heat  =  925  --  50  =  875  B.  t.  u. 

(d)  Work  done  during  adiabatic  expansion  =  1162  —  875 

=  287  B.  t.  u.  or  223300  ft.  Ibs. 

12.  Plot  to  scale,  on  a  pressure  volume  diagram,  the  line  rep- 
resenting the  adiabatic  expansion  for  the  previous  problem. 


Pressure  Vol 

For  a  pout 
expanding  Adia 
Pressure  of  100 
Superheat  to  a  f 
1.4"Hg. 

Lime  Diagram 
id  of  Steam 
batically  from  a 
#/C/'and  145° 
nal  Pressure  of 

> 
S 

^ 

25       50       75       100     125      150     175      200     225      250     275     300     325      350     375    400 

Volume,  cu.  ft. 

S8( 

I" 

Jg  M 

£  50 

3 


FIG.  3. 

Solution:  This  curve  is  easily  obtained  by  reading  the  volumes 
determined  by  the  constant  entropy  line,  1 . 69,  cutting  the  various 
pressure  lines,  assumed  as  desired.  Thus: 

Plate  used  Pressure  Volume 

2a  100                                     5.45 

2b  60                                     8.00 

2b  40                                   10.85 


52  PROBLEMS 

Plate  used  Pressure  Volume 

3b  20  19.5 

3b  10  35.9 

4b  5  66.4 

4b  2.6  119.3 

5b  1.2  237. 

5b  1.4"Hg.  397. 

13.  A  closed  metallic  tank  having  a  cubical  content  of  30  cubic 
feet  contains  3  pounds  of  steam  having  an  absolute  pressure  of 
75  pounds  per  square  inch. 

If  heat  is  now  abstracted  from  this  tank  by  pouring  cold  water 
over  it  until  the  pressure  of  the  steam  within  has  fallen  to  40  pounds 
per  square  inch  absolute,  find: 

(a)  The  initial  superheat. 

(b)  The  final  quality. 

(c)  The  heat  abstracted. 

30 
Solution:  Specific  Volume   =  -5-  =  10  cu.  ft.  per  Ib. 

o 

Hence  by  turning  to  Plate  2a,  and  running  along  the  75-pound 
line  until  it  intersects  the  10  cubic  feet  specific-volume  line,  we 
find: 

(a)  Initial  superheat  =  490°  F. 

Initial  total  heat  =  1421  B.  t.  u.  per  Ib. 

(b)  This  must  be  a  constant  volume  abstraction  of  heat  since 
the  steam  is  confined  in  a  closed  tank.     Hence  for  a  specific  volume 
of  10  cubic  feet  per  pound  and  an  absolute  pressure  of  40  pounds 
per  square  inch  we  may  read  from  Plate  2b: 

Final  quality       =95.3% 

Final  total  heat  =  1126  B.  t.  u.  per  Ib. 

(c)  The  heat  abstracted  during  any  constant-volume  process 
is  equal  to  the  loss  of  intrinsic  heat,  since  no  work  is  done  during 
such  a  process. 

The  intrinsic  heat  for  any  state  may  be  obtained  by  subtracting 
the  external  work,  as  obtained  from  Plate  8,  from  the  total  heat. 


PROBLEMS  53 

At  the  intersection  of  the  10  cubic  feet  specific- volume  line 
and  the  490°  superheat  line  on  Plate  8a,  we  read: 

The  constant  pressure  external  work  =  138.5  B.  t.  u.  per  Ib. 

Hence 

Initial  intrinsic  heat  =  1421-138.5  =  1282.5  B.  t.  u.  per  Ib. 

From  the  same  plate,  at  the  intersection  of  the  40-pound  pressure 
line  and  the  10  cubic  feet  specific-volume  line,  we  read: 

The  constant  pressure  external  work  =  73 . 8  B.  t.  u.  per  Ib. 
Hence  final  intrinsic  energy  must  be 

1126  -  73.8  =  1052.2  B.  t.  u.  per  Ib. 

Therefore  for  the  3  pounds  of  steam  cooled  at  constant  volume, 
we  have: 

Heat  abstracted  =  3  [1282 . 5  - 1052 . 2] 

=  3  [230.3]  =  690. 9  B.  t.  u. 

14.  A  pound  of  steam  having  a  pressure  of  62  pounds  per  square 
inch  absolute  and  a  temperature  of  340°  F.  is  heated  isothermally 
until  its  pressure  becomes  20  pounds  per  square  inch  absolute. 

Find: 

(a)  Initial  and  final  superheats. 

(b)  Initial  and  final  entropies. 

(c)  Initial  and  final  total  heats. 

(d)  Initial  and  final  volumes. 

(e)  Initial  and  final  intrinsic  heats. 

(f)  Heat  required  to  effect  the  change. 

(g)  Work  done  during  this  expansion. 

Solution:  From  Plates  2b  and  3b  respectively,  we  may  read: 

Temperature  of  vaporization  for  62  Ibs.  =  295°  F. 
Temperature  of  vaporization  for  20  Ibs.  =  228°  F. 

Therefore 

(a)  Initial  superheat  =  340  -  295  =    45°  F. 
Final  superheat    =  340  -  228  =  112°F. 

Then  from  Plate  2b,  for  the  pressure  of  62  pounds  and  45° 
superheat  we  may  read: 


54  PROBLEMS 

(bi)  Initial  entropy      =  1 . 67 

(ci)  Initial  total  heat  =  1201  B.  t.  u. 

(di)  Initial  volume      =7.46  cu.  ft. 

Also  from  Plate  3a,  for  the  pressure  of  20  pounds  and  112° 
superheat  we  may  read: 

(b2)  Final  entropy       =  1.803 

(c2)  Final  total  heat    =  1209.3  B.  t.  u. 

(d2)  Final  volume        =23.7  cu.  ft. 

From  Plate  8a,  for  the  pressure  of  62  pounds  and  the  superheat 
of  45°  we  read: 

The  constant  pressure  external  work  =  85  B.  t.  u. 

From  Plate  8b,  for  the  pressure  of  20  pounds  and  the  super- 
heat of  112°  we  read: 

The  constant  pressure  external  work  =  87 . 3  B.  t.  u. 
Then  we  may  obtain: 
(e)  Initial  intrinsic  heat  =  1201      -  85      =  1116  B.  t.  u. 

Final  intrinsic  heat    =  1209.3  -  87.3  =  1122  B.  t.  u. 

The  heat  required  to  go  along  a  constant  temperature  line  is 
equal  to  the  change  hi  entropy  multiplied  by  the  absolute  tem- 
perature. 

Hence  for  this  case 

(f)  Heat  required  =  (1.803  -  1.67)  (340  +  460) 

=  106.4  B.  t.  u.  given  to  the  steam. 

Representing  the  initial  and  final  states  by  the  subscripts  1  and 
2,  we  then  have,  from  the  fundamental  equation  of  thermodynamics: 


(g) 


Work 


done 


Heat       I2 


supplied  J 


"Gam  in 


intrinsic  heat 


=  106.4  -     [1122  -  1116] 
=  106.4  -  6 
=  100. 4  B.  t.  u. 

NOTE. — This  problem  is  of  much  more  importance  as  a  drill  in  thermodynamics  than  as  a  practical 
question;  because  it  is  very  difficult  to  arrange  the  necessary  apparatus  to  permit  steam  to  expand  isother- 
mally  while  in  the  superheated  field. 

Those  who  think  that  the  heat  required  to  effect  this  change  should  be  equal  to  the  difference  in  total 
heats  are  referred  to  the  meaning  of  that  term  as  given  in  the  introduction. 


PROBLEMS  55 

15.  The  pressure  in  a  steam  pipe  is  105  pounds  per  square  inch 
absolute,  and  hi  a  throttling  calorimeter,  which  is  connected  to 
this  pipe,  the  pressure  is  15  pounds  per  square  inch  absolute.    If 
the  temperature  hi  the  calorimeter  is  238°  F.,  find  the  quality  of 
the  steam. 

/ 

Solution:  From  Plate  3b,  for  a  pressure  of  15  pounds  we 
read: 

Temperature  of  vaporization  =  213°  F. 

Hence 

Superheat  in  calorimeter  =  238  -  213  =  25°  F. 
Then  from  the  opposite  page,  Plate  3a,  for  a  pressure  of  15 
pounds  and  25°  superheat  we  read: 

Total  heat  =  1163  B.  t.  u. 

The  throttling  in  a  calorimeter  takes  place  without  any  ap- 
preciable loss  of  heat  by  conduction  or  radiation,  so  the  total 
heat  remains  constant. 

Hence  by  running  along  this  total  heat  line  of  1163  B.  t.  u. 
until  we  intersect  the  105-pound  pressure  line,  we  may  read,  from 
Plate  2b: 

Quality  =  97.3% 

16.  If  the  temperature  in  the  calorimeter  of  the  previous  prob- 
lem had  been  276°  F.,  and  the  pressure  in  the  calorimeter  had  been 
16  pounds  per  square  inch  absolute,  find  the  quality  of  the  steam 
in  the  main  having  a  pressure  of  187  pounds  per  square  inch  ab- 
solute. 

Solution:  Proceeding  as  before,  from  Plate  3b  we  find  the 
superheat  hi  the  calorimeter  to  be 

276  -  216.3  =  59. 7° F. 

Then,  from  Plate  3a,  for  the  16  pound  line  and  59 . 7°  superheat 
we  read: 

Total  heat  =  1180  B.  t.  u. 

From  Plate  Ib,  at  the  intersection  of  the  1 180- total-heat  line 


56  PROBLEMS 

and  the  187-pound-pressure  line,  which  is  readily  found  by  the 
eye  between  the  185-  and  190-pound  lines,  we  read: 

Quality  =  98% 

17.  The  pressure  in  the  seventh  stage  of  a  twelve-stage  turbine 
is  10. 5  pounds  per  square  inch  absolute,  and  its  quality  is  estimated 
to  be  94%.     Supposing  that  a  fair  sample  of  the  steam  hi  this 
stage  might  be  obtained,  what  pressure  would  have  to  be  main- 
tained in  a  throttling  calorimeter  in  order  that  it  might  be  used  to 
determine  this  quality,  provided  that  there  shall  be  at  least  10° 
superheat  in  the  calorimeter? 

Solution:    From   Plate  3b,  for  an  absolute  pressure  of   10.5 
pounds  and  a  quality  of  94%  we  read : 

Total  heat  =  1085  B.  t.  u. 

Then,  running  along   this   total-heat   line   until   we   intersect 
the  10°  superheat  line,  we  read  from  Plate  6a: 

Pressure  in  calorimeter  =  0. 16  Ibs.  per  sq.  in.  abs. 

NOTE. — This  extremely  low  pressure  required  in  the  calorimeter  is  very  difficult  to  obtain  and  the 
scheme  is  not  therefore  a  very  practical  one. 

18.  If  steam  having  a  quality  of  99%  is  generated  in  an  auto- 
mobile boiler  at  a  pressure  of  400  pounds  per  square  inch  absolute, 
and  is  then  throttled  down  to  an  absolute  pressure  of  100  pounds 
per  square  inch,  find: 

(a)  Final  superheat. 

(b)  Drop  in  temperature  due  to  throttling. 

(c)  Increase  in  volume  due  to  throttling. 

Solution:    From  Plate  Ib,  for  the  pressure  of  400  pounds  and 
the  quality  of  99%  we  read : 

Initial  total  heat  =  1200  B.  t.  u.  per  pound 

Initial  temperature       =  445°  F.  (almost) 

Initial  specific  volume  =  1 . 16  cubic  feet  per  pound 

Then  from  Plate  2b,  running  along   the  1200-total-heat   line 
until  we  intersect  the  100-pound  line,  we  read: 

I 


PROBLEMS  57 

(a)  Final  superheat  =  24°  F. 

Final  specific  volume  =  4.6  cubic  feet  per  pound 
Temperature  of  vaporization  for  100  pounds  =  328°  (almost) 
Therefore  final  temperature  =  328  +  24  =  352°  F. 

(b)  The  drop  in  temperature  due  to  throttling  then  is 

445  -  352  =  93°  F. 

(c)  The  increase  in  specific  volume  due  to  throttling  is 

4.6  -  1.16  =  3.44cu.  ft.  per  Ib. 

19.  The  test  on  a  boiler  gave  the  following  data: 

Absolute  pressure,  pounds  per  square  inch 185 

Superheat  of  steam,  °F 125 

Temperature  of  feed  water,  °F 200 

Water  evaporated  per  pound  of  coal,  pounds. 9.8 

Calorific  value  of  the  coal,  B.  t.  u.,  per  pound 14300 

Find  the  efficiency  of  the  boiler  and  furnace  combined. 

Solution:    For  any  temperature  in  the  neighborhood  of  200°  F. 
the  heat  of  the  liquid  may  be  accurately  obtained  by  subtracting 
32  from  the  given  temperature,  so  in  this  case  the  heat  of  the  liquid  is 
200  -  32  =  168  B.  t.  u. 

This  result  may  also  be  obtained  from  Plate  3b  from  the  heat 
of  the  liquid  curve  for  the  temperature  of  200°  F. 

From  Plate  Ib,  for  the  pressure  of  185  pounds  and  125°  F. 
superheat  we  read: 

Total  heat  =  1268  B.  t.  u. 

Hence  the  efficiency  of  boiler  and  furnace  combined  is 

(1268  -  168)  9.8 
14300 

20.  For  each  pound  of  fuel,  a  certain  boiler  makes  10  pounds 
of  steam  having  a  quality  of  98%  and  an  absolute  pressure  of  120 
pounds  per  square  inch. 

If  the  temperature  of  the  feed  water  is  80°  F.,  find: 

(a)  Factor  of  evaporation. 

(b)  Equivalent  evaporation. 


58  .      PEOBLEMS 

Solution:    The  heat  of  the  liquid  at  feed  temperature  is 

80  -  32  =  48  B.  t.  u. 

From  Plate  Ib,  for  the  pressure  of  120  pounds  and  a  quality  of 
98%  we  read: 

Total  heat  =  1172  B.  t.  u. 
Then,  by  definition,  we  have: 

Heat  absorbed  per  Ib.  of  steam 

(a)  Factor  of  evaporation  =  ^—    ^rr — .    f  .  OIOOT^ 

Latent  heat  of  steam  at  212°  F. 

1172  -  48 
970.4 

(b)  Equivalent  evaporation  is  the  amount  of  water  that  would 
be  evaporated  from  and  at  212°  F.  by  the  same  amount  of  heat  as  is 
actually  absorbed  per  pound  of  fuel.     The  equivalent  evaporation 
in  this  case  will  therefore  be 

10  X  1.159  =  11. 59  pounds 

21.  If,  for  each  pound  of  fuel,  the  boiler  of  the  previous  problem 
had  delivered  9.5  pounds  of  steam  having  a  superheat  of  175° 
and  at  the  same  pressure  as  before,  find  the  factor  of  evaporation 
and  equivalent  evaporation. 

Solution:    From  previous  problem 

Heat  of  the  liquid  =  48  B.  t.  u. 

From  Plate  2a,  for  pressure  of  120  pounds  and  a  superheat 
of  175°  we  read: 

Total  heat  =  1282  B.  t.  u. 
Then,  as  before, 

12§2  48 

(a)  Factor  of  evaporation  =  — 070"! —  =  1.273 

(b)  Equivalent  evaporation  =  9.5  X  1.273  =  12.1  pounds. 

22.  A  steam  main  is  supplied  with  steam  from  two  boilers,  one 
of  which  furnishes  6,000  pounds  of  steam  per  hour,  the  quality 
being  98%,  while  the  other  boiler  furnishes  4,000  pounds  of  steam 
per  hour,  the  superheat  being  95°  F.    Assuming  that  both  boilers 
deliver  the  steam  at  a  pressure  of   190  pounds  per  square  inch 


PROBLEMS  59 

absolute,  and  neglecting  all  losses,  find  the  condition  of  the  steam 
in  the  main. 

Solution:  From  Plate  Ib,  for  the  pressure  of  190  pounds  and 
the  quality  of  98%  we  read: 

Total  heat  =  1180  B.  t.  u. 

From  the  same  Plate,  for  same  pressure  and  for  a  superheat  of 
95°  we  read: 

Total  heat  =  1253  B.  t.  u. 

Then  for  the  mixture,  we  have: 

1180  X  6000  +  1253  X  4000       10An  _ 
Total  heat  =  -  60Q0  +  4QoQ  =  1209  B.  t.  u. 

Hence,  for  the  pressure  of  190  pounds  and  a  total  heat  of  1209, 
from  Plate  Ib  we  read: 

Superheat  in  main  =  18°  F. 

23.  Assuming  no  loss  by  radiation,  conduction,  or  leakage,  what 
would  have  been  the  condition  of  the  steam  in  the  steam  main 
of  the  previous  problem,  had  there  been  a  loss  of  pressure  due  to 
friction  hi  the  pipes,  so  that  the  pressure  in  the  mam  was  only 
175  pounds  per  square  inch  absolute? 

Solution:  Since  there  is  no  loss  of  heat,  the  total  heat  of  the 
mixture  must  be  the  same  as  in  the  previous  problem.  From 
Plate  Ib,  we  may  follow  the  1209-total-heat  line  until  it  intersects 
the  175-pound  line,  where  we  read: 

Superheat  in  the  main  =  20°  F. 

2"4.  Two  boilers,  A  and  B,  are  connected  to  the  same  steam  main 
and  the  following  observations  were  made: 

Total  steam  passing  through  the  main  =  12000  Ibs.  per  hr. 

Average  pressure  in  the  main  =  195  Ibs.  per  sq.  in.  abs. 

Average  pressure  leaving  boiler  A  =  200  Ibs.  per  sq.  in.  abs. 

Average  pressure  leaving  boiler  B  =  210  Ibs.  per  sq.  in.  abs. 

Average  superheat  in  the  main  =  56°  F. 

Average  superheat  leaving  boiler  A  =  33°  F. 

Average  superheat  leaving  boiler  B  =  84°  F. 


60  PROBLEMS 

Find  the  weight  of  steam  coming  from  each  boiler,  assuming 
no  losses  by  radiation  or  conduction. 

Solution:     From  Plate  Ib,  we  read: 

For  steam  from  boiler  A,  having  the  pressure  of  200  pounds 
and  superheat  of  33°, 

Total  heat  =  1220  B.  t.  u. 

For  steam  from  boiler  B,  having  the  pressure  of  210  pounds  and 
superheat  of  84°, 

Total  heat  =  1250  B.  t.  u. 

For  steam  in  the  main,  having  the  pressure  of  195  pounds  and 
superheat  of  56°, 

Total  heat  =  1232.5  B.  t.  u. 

Let  WA  =  Ibs.  of  steam  per  hr.  coming  from  boiler  A 

Let  WB  =  Ibs.  of  steam  per  hr.  coming  from  boiler  B 

Let  HA  =  the  total  heat  of  steam  B.  t.  u.  per  Ib.  from  boiler  A 

Let  H#  =  the  total  heat  of  steam  B.  t.  u.  per  Ib.  from  boiler  B 

Let  Hm  =  the  total  heat  of  steam  B.  t.  u.  per  Ib.  in  main. 

Since  there  has  been  no  loss  of  heat,  we  may  write : 
WAHA  +  WB&B  =  (WA  +  WB)  Hw 
From  the  conditions  of  the  problem 

WB  =  12000  -  WA 
Combining  these  two  equations  we  have: 

WA  HA  +  [12000  -  WA]  H5  =  12000  Hm 
.'.  [HA  -  HB]  WA  =  12000  [Hm  -       ~ 


•      TTTT  ion™ 

..WA=  12000 


.  10000  .     "  1250-1 

12000 


=  7000  pounds  per  hr. 
Hence  WB  =  5000  pounds  per  hr. 


PROBLEMS  61 

25.  The  exhaust  steam  from  an  engine  is  delivered  to  the  heating 
system  of  a  building.     If  the  pressure  in  this  system  is  maintained 
constant  at  17  pounds  per  square  inch  absolute,  and  if  the  quality 
of  the  steam  exhausted  by  the  engine  is  88%,  find  the  amount  of 
heat  that  would  be  given  up  by  this  system  when  the  engine  is 
using  3,000  pounds  of  steam  per  hour.    The  water  in  the  return 
pipes  is  delivered  to  the  boiler  room  at  a  temperature  of  210°  F. 

Solution:     From  Plate  3b,  for  the  pressure  of  17  pounds  and 
quality  of  88%  we  read: 

Total  heat  =  1037  B.  t.  u. 

The  heat  returned  to  the  boiler  room  is  equal  to 

Heat  of  the  liquid  =  210  -  32  =  172  B.  t.  u. 
The  heat  given  up  by  the  total  weight  of  exhaust  steam  therefore  is 

3000  (1037  -     187)  =  2577000  B.  t.  u.  per  hr. 

26.  A  12,000-horse-power  steam  turbine  required  9.6  pounds  of 
steam  per  horse-power  hour  when  supplied  with  steam  having  a 
pressure  of  175  pounds  per  square  inch  absolute  and  132  degrees 
of  superheat.     If  the  exhaust  pressure  were  2  inches  of  mercury 
absolute,  find 

1  (a)  Heat  supplied  per  h.p.  hr. 

(b)  Delivered  thermal  efficiency. 

(c)  Cycle  efficiency. 

(d)  Theoretical  water  rate. 

(e)  Efficiency  ratio. 

Solution :    Let  dbcde,  Fig.  4,  represent  the  cycle  upon  which  the 
ideal  turbine   operates. 

Heat  supplied  per  pound  of  steam 

=  area  mabcdn 

=  Total  heat]d  —  Heat  of  the  liquid]a 

=  1270  -69  =  1201  B.  t.  u. 

This  value  of  total  heat  at  the  point  d  is  read  directly  from 
Plate  Ib  for  the  pressure  of  175  pounds  and  superheat  of  132°. 


62 


PROBLEMS 


The  heat  of  the  liquid  for  the  point  a  is  read  from  Plate  5b,  at 
the  intersection  of  the  heat  of  the  liquid  curve  and  the  2  inches  of 
mercury  pressure  line. 


Entropy 

FIG.  4. 

Then,  since  the  turbine  requires  9 . 6  pounds  of  steam  per  h.p. 
hour, 

(a)  Heat  supplied  per  h.p.  hr.  =  9.6  X  1201  =  11529.6  B.t.u. 

„  N  T.  ,.  One  horse-power  hour  in  B.t.u. 

(b)  Delivered  thermal  efficiency  =  — ^ — 7 —    r   i r r 

Heat  supplied  per  h.p.  hr. 

2545 


(c)  Cycle  efficiency 


11529.6  " 
Area  abcde 


Area  mabcdn 
Total  heat]d  -  Total  heat]e       1270  -  915 

1201  1201 

355 


1201 


=  29.55% 


This  value  of  total  heat  for  the  point  e  is  obtained  from  Plate 
5b,  by  reading  the  total  heat  at  the  intersection  of  the  pressure  line 


PROBLEMS  63 

for  2  inches  of  mercury  and  the  1  .  64  entropy  line.    This  entropy  is 
obtained  from  Plate  Ib,  for  the  state  d. 

2545 
(d)  Theoretical  water  rate  =  Net  work  per  lb.  of  steam 

2545       ,_  _  , 

=7.17  pounds  per  h.p.  hr. 


,  N  __.  .  Actual  thermal  efficiency 

(e)  Efficiency  ratio  Cycle  efficiency 


r,~  .  ..         Theoretical  water  rate 

or  Efficiency  ratio  =  —  r—  -,  —  ;  -  7  -  7  — 

Actual  water  rate 


27.  Suppose  that  all  the  conditions  of  operation  remain  the 
same  as  hi  the  previous  problem,  except  the  water  rate  and  the 
vacuum.  If,  by  reducing  the  exhaust  pressure  to  1  inch  of  mercury 
absolute,  the  water  rate  was  actually  reduced  to  9  pounds  per  h.p. 
hour,  find  (a)  (b)  (c)  (d)  and  (e)  as  before. 

Solution:    Referring  to  the  same  figure  modified  by  the  line 
a'e'  and  proceeding  in  the  same  manner,  we  have: 
From  Plate  5b, 

Heat  of  the  liquid]a'=  47  B.  t.  u. 

(a)  Heat  supplied  per  hr.  =  9  [1270  -  47]  =  11  007  B.  t.  u. 

2545 

(b)  Delivered  thermal  efficiency  =  777^7  =  23  .  1% 

From  Plate  6b 

Total  heat]e'  =  882  B.  t.  u. 
.'.    Net  work  per  cycle   =  1270  -  882  =  388  B.  t.  u. 

Hence 

388 

(c)  Cycle   efficiency  =  127Q  _  47  =31.7% 

2545 

(d)  Theoretical  water  rate  =  -™-  =6.56  pounds  per  h.p.  hr, 

231 

(e)  Efficiency  ratio  =  =  72. 


64  PROBLEMS 

28.  A  steam  turbine  receives  steam  at  a  pressure  of  80  pounds 
per  square  inch  absolute  and  a  superheat  of  77°  F.     If  this  steam 
expands  in  a  single  set  of  nozzles  to  a  condenser  pressure  of  1 . 4  Ibs. 
per  square  inch  absolute,  and  if  we  assume  no  nozzle  losses,  find, 
for  the  steam  leaving  nozzles: 

(a)  Quality. 

(b)  Specific  volume. 

(c)  Velocity. 

Solution:    From  Plate  2b,  for  the  initial  state  we  read: 
Total  heat  =  1223  B.  t.  u. 
Entropy      =1.67 

From  Plate  5b,  for  the  final  pressure  and  the  entropy  1 . 67,  we 
read: 

Total  heat  =  950  B.  t.  u. 

(a)  Quality  =  84.4% 

'  (b)  Specific  volume  =  204  cu.  ft.  per  Ib. 

The  energy  which  has  been  transformed  into  velocity  is  equal 
to  the  difference  in  total  heats. 

Initial  total  heat  -  Final  total  heat  =  1223  -  950  =  273  B.  t.  u. 
per  Ib. 

Then,  from  Table  IV,  we  read: 

(c)  Velocity  =  3696  feet  per  second. 

29.  The  theoretical  steam  engine  operates  on  the  incomplete 
expansion  cycle,  as  shown  by  abcde,  Figs.  5  and  6.      The  pres- 
sure at  the  throttle  is  165  pounds  per  square  inch  absolute  and  the 
superheat  is  143°  F.     If  the  back  pressure  is  2 . 8  pounds  per  square 
inch  absolute  and  the  expansion  ratio  is  6,  find,  for  the  ideal  cycle: 

(a)  Pressure  at  release. 

(b)  Quality  at  release. 

(c)  Net  work  of  cycle  in  B.  t.  u. 

(d)  Cycle  efficiency. 

(e)  Water  rate  of  an  ideal  engine  working  on  this  cycle. 


PROBLEMS 


65 


Volume 

FIG.    5. 


Entropy 

FIG.  6. 

Solution:    From  Plate  Ib,  for  the  pressure  of  165  pounds  and 
the  superheat  of  143°  we  read: 

Initial  total  heat  =  1274  B.  t.  u.  per  Ib. 

Initial  specific  volume  =  3 . 4  cu.  f t.  per  Ib. 
Initial  entropy  =  1 . 65 


66  PROBLEMS 

Since  the  expansion  ratio  is  6,  the  specific  volume  at  release 
must  be 

Vc  =  6  Vb  =  6  X  3.4  =  20.4  cu.  ft.  per  Ib. 

For  the  ideal  cycle,  the  expansion  line  is  an  adiabatic.  Hence 
by  following  the  1.65  entropy  line  until  it  intersects  the  20.4 
specific- volume  line,  we  may  read  from  Plate  3b : 

(a)  Release  pressure  =  18.4  Ibs.  per  sq.  in.  abs. 

(b)  Release  quality    =93.7% 

Total  heat]c  =  1094.5  B.  t.  u. 

In  order  to  find  the  net  work  of  the  cycle  we  may  proceed  in 
several  ways,  one  being  to  subtract  the  heat  rejected  from  the  heat 
supplied.  Referring  to  the  temperature  entropy  diagram,  Fig.  6, 
we  have: 

Heat  supplied  =  area  meabn 

=  Total  heat]&  —  Heat  of  the  liquid]  e 
=  1274  -  107  =  1167  B.  t.  u.  per  Ib. 

The  heat  of  the  liquid  at  the  point  e  is  obtained  from  Plate  4b 
for  the  2 . 8  pounds  back-pressure  line. 
Heat  rejected   =  area  cdemn. 

Intrinsic  ~]          Intrinsic  ~i          Work~ie 
"  heat        Jc        heat        Je        done  Jc 
=  [1094.5  -  69]  -  107  +  11=  929.5  B.  t.  u.  per  Ib. 

This  intrinsic  heat  at  c  is  obtained  by  subtracting  from  the 
total  heat  for  this  point  the  constant  pressure  external  work  as  ob- 
tained from  Plate  8b.  We  know  the  specific  volume  at  c  to  be 
20.4  cubic  feet  per  pound,  and  the  quality  at  c  has  already  been 
found  to  be  93.7%.  Therefore  we  may  read  from  this  plate  the 
value  of  this  external  work,  69  B.  t.  u.,  as  above  given. 

The  intrinsic  heat  at  e  is  merely  the  heat  of  the  liquid  for  the 
exhaust  pressure  and  has  already  been  found  to  be  107  B.  t.  u. 

The  work  done  in  going  from  c  to  e  is  the  same  as  that  done  in 
going  from  d  to  e  since  the  work  done  from  e  to  d  is  zero,  the  volume 
being  constant.  But  the  work  from  d  to  e  may  be  obtained  directly 
from  Plate  8b.  Thus  by  running  down  the  constant  specific- volume 


PROBLEMS  67 

line,  20.4,  until  it  intersects  the  2.8  pound  back-pressure  line,  we 
may  read: 

Work  doneT  =  11  B.  t.  u.  per  Ib. 

Jo 

Then  from  the  above  values,  we  have: 

(c)  Net  work  of  cycle  =  Heat  supplied  —  Heat  rejected 

=  1167  -  929.5  =  237.5  B.  t.  u.  per  Ib. 

Net  work  of  cycle       237.5  Ir~ 

(d)  Cycle  effidency      =      Heat  supplied      =  Tl67    =  20'35% 

2545  2545 

(e)  Ideal  water  rate       =  Net  work  per  Ib.  of  steam       =  2^1 

=  10.71  Ibs.  per  h.p.  hr. 

NOTE.  —  For  the  cycle  abcde,  Fig.  5  would  commonly  be  drawn  with  a  different  ratio  of  scales  so  that 
the  distance  representing  the  admission  pressure  would  be  nearly  the  same  as  the  distance  representing 
the  volume  at  d.  In  this  case,  however,  it  was  desired  to  draw  to  scale  the  entire  diagram,  including 
the  "toe"  cfd,  so  that  is  why  the  volume  at  d  may  seem  to  be  entirely  too  small  for  any  real  engine  having 
a  ratio  of  expansion  equal  to  six. 

Since  a  pound  of  water  occupies  a  space  of  only  about  .  016  cubic  feet,  it  is  evidently  impossible  in  Fig. 
5  to  represent  such  a  volume  with  any  line  other  than  the  zero  volume  line  ea.  Even  with  a  scale  of  volumes 
as  large  as  that  used  for  Fig.  7  with  Problem  34,  the  volume  of  the  water  can  scarcely  be  shown. 

30.  Find  the  net  work  of  cycle  in  the  previous  problem  by  the 
method  of  combining  the  two  cycles  dbce'  and  e'cde,  Figs.  5  and  6. 

Solution: 

Net  work  of]        m  ,  ,-,         m  ,  in 

cvcle  -  Totall     -  Totall 

heatJ*        heatJc 


=  1274  -  1094.5 

=  179.5B.  t.  u.  perlb. 

These  values  are  obtained  from  Plates  Ib  and  4b  as  before. 
Net  work  of 


cycle 
e'cde 


c  e 

=  Work     -  Work 

J  -* 


=  69  -  11  =  58  B.  t.  u.  per  Ib. 

These  values  may  be  obtained  from  Plate  8b  as  before. 
Hence  the  net  work  of  the  cycle  abcde  is 

179.5  +  58  =  237.5  B.  t.  u.  per  Ib. 


68  PROBLEMS 

31.  For  the  previous  problem  find  the  amount  of  work  lost  due 
to  incomplete  expansion  and  thereby  check  the  result  of  preceding 
problem. 

Solution:  The  work  lost  due  to  incomplete  expansion  is  rep- 
resented by  cfd  in  Fig.  5  or  6. 

By  inspection  of  these  figures  it  will  be  seen  that 
cfd  =  e'cfe  —  e'cde 

Work       j  Total!        Total"!)      j  w    ,  f     —    ,  -i« ) 
°r     lost  jheat  I  ~  heat  J,}  ~  \  Work I"  WorkJd[ 

=  1094.5  -  977.3  -69  +  11 
=      59.2B.  t.  u.  perlb. 

The  total  heat  at  c  is  obtained  from  Plate  3b,  as  before.  The 
total  heat  at  /  is  found  from  Plate  4b,  for  the  given  back  pressure 
of  2.8  pounds  and  the  1.65  entropy  line. 


]c 
is  found  from  Plate  8b  as  before. 
e 

Work       is  found  from  Plate  8b  as  before. 

_Ja 


To  check  the  net  work  of  the  incomplete  expansion  cycle  we  may 
subtract  the  "toe"  from  the  complete  expansion  cycle.  Thus 
abode  =  abfe  —  cfd 

v      ,,  Total"!        Total"! 

But  the  cycle  abfe  =   ,  —  , 

heat  J6        heat  J/ 

=  1274  -  977.3 

=  296.7B.  t.  u.  perlb. 

Hence  the  cycle  abcde  is  equal  to 

296.7  -  59.2  =  237.5  B.  t.  u.  per  Ib. 

This  checks  the  results  of  29  and  30. 

32.  Find  the  ratio  of  volumes  of  the  cylinders  necessary  for  the 
complete  and  incomplete  expansion  cycles  of  problem  29,  and  also 
find  the  per  cent,  reduction  in  the  net  work  due  to  the  incomplete 
expansion. 


PROBLEMS  69 

Solution:    From  problem  31  we  have: 
Per  cent,  reduction  in  net  work  is 

^  -  19  9% 
296.7 

From  problem  29  the  specific  volume  at  e  was  20.4  cubic  feet 
per  pound. 

From  Plate  4b,  for  the  given  back  pressure  of  2.8  pounds  and 
the  entropy  1.65,  we  obtain 

Specific  volume  at  /  =  108  .  2  cu.  ft.  per  Ib. 

Then,  since  the  cylinder  volumes  for  any  case  would  be  propor- 
tional to  the  specific  volumes  as  thus  found  for  the  ideal  cycle,  we 

have. 

Volume  for  complete  expansion         108.2 
Volume  for  incomplete  expansion  "'   20.4 

33.  Find  the  per  cent,  increase  in  the  net  work  of  the  cycle  of 
problem  30  for  a  back  pressure  of  1  pound  per  square  inch  absolute 
instead  of  2.8  pounds. 

Solution:  By  referring  to  the  solution  of  problem  30,  we  may 
see  that  by  changing  the  back  pressure  only,  the  solution  remains 
the  same  except  the  value  of  the  work  done  during  exhaust.  From 
Plate  8b,  running  down  the  specific  volume  line  20.4,  which  is  the 
specific  volume  at  release,  until  we  intersect  the  1  pound  back- 
pressure line,  we  obtain 

Work!  '   =  3.8  B.  t.  u.  per  Ib. 

-Id 

In  problem  30  the  work  done  against  the  exhaust  pressure  was 
11  B.  t.  u.  per  pound.     This  is  an  increase  of  7.2  B.  t.  u.  per  pound. 
Hence 
Per  cent,  increase  in  the  net  work  of  the  cycle  is 


NOTE.  —  In  an  actual  engine  operating  under  these  conditions  much  less  gain  than  this  would  be  realized 
on  account  of  greater  loss  due  to  cylinder  condensation,  more  work  required  for  vacuum  pump,  and  more 
heat  required  to  heat  the  feed  water. 

34.  A  direct-acting  steam  pump  operates  on  a  non-expansive 
cycle  as  shown  by  abed,  Fig.  7. 


70  PROBLEMS 

The  engine  is  supplied  with  steam  having  a  pressure  of  120 
pounds  per  square  inch  absolute  and  a  quality  of  97%.    If  the  ex- 
haust pressure  is   18  pounds  per  square 
inch  absolute,   find  the  net  work  of  the 
f  cycle,  and  the  ideal  water  rate. 


Solution:  From  Plate  8a,  for  the  in- 
itial pressure  of  120  pounds  and  quality 
of  97%  we  find: 

Workl    =  80  B.  t.  u.  per  Ib. 

Ja 

Specific  volume  at  b  =  3 . 6  cu.  ft.  per  Ib. 

Then,  running  down  this  constant  vol- 
ume line  until  we  intersect  the  back  pres- 
sure of  18  pounds,  we  have: 

Workl     =  12  B.  t.  u.  per  Ib. 

Then  the  net  work  of  the  cycle  becomes 

80  -  12  =  68  B.  t.  u.  per  Ib. 
and  the  ideal  water  rate  is  therefore 
2545 


68 


=  37.4  Ibs.  per  h.p.  hr. 


35.  In  a  uni-flow  locomotive  the  clearance  is  16%  and  com- 
pression takes  place  during  90%  of  the  stroke. 

If  the  back  pressure  is  16  pounds  per  square  inch  absolute  and 
the  quality  of  the  steam  at  the  beginning  of  compression  is  96%, 
find  the  condition  of  the  steam  at  the  end  of  the  stroke,  assuming 
the  compression  to  be  adiabatic. 

Solution:  From  Plate  3b,  for  the  pressure  of  16  pounds  and 
quality  of  96%  we  may  read: 

Entropy  =  1.6925 

Initial  specific  volume  =  23 . 8  cu.  ft.  per  Ib. 


PROBLEMS  71 

We  must  next  find  the  specific  volume  at  the  end  of  compression, 
which  is  equal  to  the  specific  volume  at  the  beginning  of  com- 
pression divided  by  the  ratio  of  compression. 

Ratio  of  compression  =  -  '       =  6 . 63 

Hence 

OQ    Q 

Final  specific  volume  =  ^-^  =  3 . 59  cu.  ft.  per  Ib. 

O.  Do 

Then  by  going  along  the  entropy  line  1 . 6925,  until  we  intersect 
this  specific  volume  line  3 . 59,  we  read  from  Plate  la : 

Final  pressure     =  175  Ibs.  per  sq.  in.  abs. 
Final  superheat  =  240°  F. 

36.  An  engine  having  a  steam  jacket  has  a  clearance  of  15% 
of  the  piston  displacement  which  is  3 . 17  cubic  feet.     It  is  found 
from  test  results  that  the  weight  of  steam  inside  the  cylinder  dur- 
ing expansion  is  0.163  pounds.     If  release  occurs  at  95%  of  the 
stroke  at  a  pressure  of  19  pounds  per  square  inch  absolute,  find  the 
condition  of  the  steam  at  release. 

Solution:  Actual  volume  of  the  steam  hi  the  cylinder  is  equal 
to  the  clearance  volume  plus  the  piston  displacement.  But  since 
the  clearance  is  given  in  terms  of  the  piston  displacement,  we  have: 

Actual  volume  =  1.10  X  3.17  =  3.49  cu.  ft. 
Specific  volume  of  the  steam  in  the  cylinder  is 

Actual  volume        3.49        M  »,          ,, 

~TT  .  ,  ,      -  =  n  ir?0  =21.4  cu.  ft.  per  Ib. 
Weight  0.163 

Then  from  Plate  3a,  for  the  pressure  of  19  pounds  and  a  specific 
volume  of  21 .4  we  find: 

Superheat  =  9°F. 

37.  The  temperature  in  a  condenser  is  87°  F.  and  the  absolute 
pressure  is  equal  to  1.43  inches  of  mercury.     Find  the  weight  of  air 


72  PROBLEMS 

which  must  be  removed  per  pound  of  exhaust  steam  if  the  steam 
entering  the  condenser  has  a  quality  of  85%. 

Solution:  From  the  curve  giving  temperature  of  vaporization, 
Plate  5b,  for  87°,  we  find: 

Pressure  of  wet  steam  =  1 . 29"  Hg. 
and  for  this  pressure  and  a  quality  of  85% 

Specific  volume  =  442.  cu.  ft.  per  Ib. 

The  air  pressure  is  equal  to  the  total  pressure  less  the  pressure  of 
the  steam,  that  is, 

1.43  -  1.29  =  0.14"Hg. 

1  inch  of  mercury  at  58.1°  being  equal  to  0.49  pound  per  square 
inch,  the  air  pressure  in  the  condenser  is 

P  =  .14  X  .49  X  144  Ib.  per  sq.  ft. 
The  absolute  temperature  in  the  condenser  is 
T  =  460  +  87  =  547°. 

Hence  the  weight  of  air  existing  in  the  same  space  as  each  pound 
of  steam  is 

PV         (.14  X  .49  X  144)  442 
=  RT  53.3  X  547 

38.  If  the  temperature  of  the  atmosphere  is  68.5°  F.  and  the 
relative  humidity  is  70%,  find  the  pressure  due  to  this  moisture 
and  the  weight  of  moisture  in  each  1,000  cubic  feet  of  the  atmosphere. 

Solution:  From  the  temperature  of  vaporization  curve  of 
Plate  6b,  for  the  temperature  of  68.5°  we  may  read : 

Saturation  pressure       =  0 . 1"  Hg. 

Since  the  relative  humidity  is  almost  exactly  equal  to  the 

Actual  vapor  pressure 

Saturation  pressure        '  li  follows  that 


PROBLEMS  73 

Actual  vapor  pressure  =  .70  X  0.7  =  0.49"  Hg. 

=  .49  X  .49  =  0.24  Ib.  per  sq.  in. 

Then  from  Plate  6a,  for  this  pressure,  0.24  pound  and  the  temperature 
68.5°  given  by  the  temperature  scale  on  the  right-hand  side  of  the 
sheet,  we  may  read : 

Specific  volume  =  1,310  cu.  ft.  per  Ib. 

Superheat  =  10°  F. 

It  is  interesting  to  observe  that  this  means  the  vapor  in  the  air 
in  this  condition  is  merely  superheated  steam  having  a  pressure  of 
0.24  pound  per  square  inch  absolute  or  0.49  inch  of  mercury  and 
a  superheat  of  10°.  Hence  the  weight  of  this  steam  contained 
in  1,000  cubic  feet  of  space  would  be 

1000 


1310 


=  0.7641b. 


Then  by  Dalton's  law  of  partial  pressures,  for  the  given  vapor 
pressure  and  temperature,  this  same  weight  will  be  in  this  space 
when  it  is  also  full  of  air. 

NOTE. — This  scale  of  approximate  temperatures,  placed  on  the  upper  right-hand  corner  of  Plate  6a, 
is  not  intended  to  be  sufficiently  accurate  to  solve  all  hygrometric  problems.  See  page  15  for  the 
variation  of  this  scale  from  the  true  temperatures. 


39.  Find  the  diameter  of  pipe  to  furnish  steam  for  a  9,000  kw. 
turbine,  if  it  requires  12.8  pounds  per  kw.-hr.  when  running  at  full 
load  with  a  pressure  of  200  pounds  per  square  inch  absolute  and  145° 
superheat  at  the  throttle.  Allow  a  velocity  of  8,000  feet  per 
minute. 

Solution:  From  Plate  Ib,  for  the  steam  at  the  throttle  with  a 
pressure  of  200  pounds  and  superheat  of  145°  we  find: 

Specific  volume  =  2 . 84  cu.  ft.  per  Ib. 

The  area  of  the  pipe  in  square  feet  is  equal  to 

Volume  in  cu.  ft.  per  min.     Ibs.  per  kw.-hr.  X  kw.  X  sp.  vol. 
Velocity  in  ft.  per  min.  60  X  velocity 


74  PROBLEMS 

But  if  the  diameter  of  the  pipe  in  inches  is  d,  then  the  area  in  square 
feet  is  also  equal  to 


X 

.  4  X  144  X  water  rate  X  load  X  sp.  vol. 
Hence  d  =  ^  -  ,  X  60  X  velocity 

1   748     I  wa^er  ra^e  X  load  X  sp.  vol. 
>l        velocity  in  ft.  per  min. 

,  74R     1 12.8  X  9,000X2.84 
'  M  8,000 

=  11.16". 
This  means,  of  course,  that  a  11-inch  pipe  would  be  used. 

40.  The  Parson's   turbine   at  the  Fiske  Street  Station  of  the 
Commonwealth  Edison  Co.,  Chicago,  has  an  exhaust  opening  to 
the  condenser  of  252  square  feet.    If  the  water  rate  for  a  back 
pressure  of  1  inch  of  mercury  and  a  load  of  25,000  kw.  is  11.65 
pounds  per  kw.-hr.,  find  the  velocity  through  this  opening,  assuming 
the  steam  has  a  quality  of  80%. 

Solution:    From  Plate  6b,  for  the  pressure  of  1  inch  of  mercury 
and  quality  of  80%  we  find : 

Specific  volume  of  exhaust  steam  =  523  cu.  ft.  per  Ib. 
Then 

Volume       11.65  X  25,000  X  523 
Velocity  =  "AST  =  "  60  X  252  =  10>080  ft' per  ^ 

41.  If,  in  the  second  stage  of  a  Curtis  turbine,  the  pressure  at 
the  entrance  to  the  nozzle  is  56  pounds  per  square  inch  absolute,  the 
superheat  64°,  and  the  pressure  in  the  next  stage  36.8  pounds  per 
square  inch  absolute,  find  the  cross-sectional  area  of  each  nozzle 
for  this  stage  in  order  that  112  such  nozzles  will  give  a  flow  of  96,000 
pounds  of  steam  per  hour.    Assume  the  coefficient  of  velocity  for 
this  condition  to  be  96%. 


PROBLEMS  75 

Solution :    From  Plate  2b,  for  a  pressure  of  56  pounds  and  super- 
heat of  64°  we  find: 

Total  heat          =  1,208  B.  t.  u. 
Entropy  =1.69 

Then,  by  following  this  entropy  line  until  we  intersect  the  36 . 8- 
pound  line  on  Plate  3a,  we  may  read: 

Total  heat          =  1,173  B.  t.  u. 
Specific  volume  =  11.6  cu.  ft.  per  Ib. 

Then  the  theoretically  available  energy  to  produce  velocity  in  this 
stage  is 

1,208  -  1,173  =  35  B.  t.  u.  per  Ib. 

and  from  Table  IV  this  velocity  is  1,323  feet  per  second. 

For  the  given  nozzle  coefficient  the  velocity  would  therefore  be 
.96  X  1,323  =  1,270  ft.  per  sec. 

Let  A  =  area  in  sq.  in.  of  each  nozzle. 

Let  N  =  number  of  nozzles. 

Let  v  =  velocity  in  ft.  per  sec. 

Let  V  =  specific  volume  of  steam  as  it  passes  the  throat. 

Then  the  flow  in  pounds  per  hour  is 

3,600NAv        25NAv 


F  =: 


144V 


.  A         FV  96,000  X  11.6         _Q1Q 

' ' A  =25N^  ==  25  X  112  X  1270  =          3  Sq'  m' 

For  this  and  the  following  problem  it  will  be  observed  that 
the  ratio  of  pressures  between  stages  is  considerably  over  57%, 
which  means  that  the  maximum  velocity  occurs  at  the  throat  of 
the  nozzle  and  that  the  nozzle  will  therefore  have  no  divergent 
part. 

42.  Find  the  size  of  nozzles  and  the  pressures  in  the  third  and 
fourth  stages  for  the  turbine  of  the  previous  problem  if  in  each  of 
these  stages  there  are  to  be  82  nozzles  and  the  theoretically  avail- 
able energy  is  to  be  35  B.  t.  u.  per  stage.  Assume  that  the  nozzle, 
bucket,  and  rotational  losses  amount  to  25%  of  this  energy.  Neglect 


76  PROBLEMS 

radiation  loss  and  leakage  of  steam  and  assume  nozzle  coefficient 
to  be  96%  as  before. 

Solution:  The  losses  in  each  stage  go  to  reheat  the  steam,  and 
this  reheating  is  assumed  to  take  place  at  constant  pressure  after 
the  expansion  in  the  nozzle,  since  the  nozzle  loss  is  very  small. 
Hence,  from  the  previous  problem,  the  steam  enters  the  third-stage 
nozzle  in  the  following  condition: 

Pressure  =  36.81bs. 

Total  heat          =  1,173  +  .25  X  35  =  1,181.8  B.  t.  u. 

Then,  from  Plate  3a,  for  these  two  values  we  obtain: 

Entropy  =  1.70+ 

Superheat  =  27° 

Since  it  is  desired  to  give  up  35  B.  t.  u.  in  this  stage,  we  now  follow 
down  this  entropy  line  until  the  total  heat  becomes 

1,181.8  -  35      =  1,146. 8  B.  t.  u. 
and  from  Plate  3b  we  find  for  this  total  heat  and  entropy  of  1 . 70+ 

Pressure  =23.5  Ibs. 

Specific  volume  =  17.1  cu.  ft. 

The  velocity  corresponding  to  35  B.  t.  u.  is  from  Table  IV  equal  to 
1,323  feet  per  second.  Hence,  using  the  same  notation  as  before, 
the  area  of  the  throat  of  the  nozzle  for  the  third  stage  is 

F  V  96,000  X  17.1 

=  25N^  ==  25  X82X(.96  X  1,323)  = 

Then  the  steam  enters  the  nozzle  of  the  fourth  stage  in  the  following 
condition : 

Pressure  =  23. 5  Ibs. 

Total  heat          =  1,146.8  +  .25  X  35  =  1,155.6 

From  these  two  values  and  Plate  3a  we  obtain 

Entropy  =  1.713 

Quality  =99.5% 


PROBLEMS  77 

and  by  following  this  entropy  line  until  the  total  heat  becomes 

1,155.6-35       =  1,120. 6  B.t.  u. 
we  have  from  Plate  3b 

Pressure  =14.6  Ibs. 

Specific  volume  =  26 . 2  cu.  ft. 

Hence  the  area  of  the  throat  of  the  nozzle  for  the  fourth  stage  is 

FV  96,000X26.2 

=  25  Nv  =:   25  X  82  X  (.96  X  1,323) 

s 

43.  Steam  expands  through  a  properly  shaped  divergent 
nozzle  from  an  initial  pressure  of  125  pounds  per  square  inch  absolute 
and  220°  of  superheat  to  a  final  pressure  of  1  pound  per  square  inch 
absolute.  Find  the  proper  cross-sectional  areas  of  this  nozzle  to 
permit  a  flow  of  one  pound  per  second,  assuming  adiabatic  expansion 
and  no  friction. 

Solution:  From  Plate  2a,  for  the  pressure  of  125  pounds  and 
superheat  of  220°  we  may  read: 

Total  heat  =  1,305  B.  t.  u. 

Entropy  =1.71 

and  from  Plate  5b  for  this  entropy  and  a  pressure  of  1  pound  we 
find: 

Total  heat  =  955  B.  t.  u. 

Specific  volume  =  285  cu.  ft. 

Then  the  total  available  energy  is 

1,305  -  955         =  350  B.  t.  u.  per  Ib. 

Let  the  nozzle  be  divided  into  sections,  so  that  in  each  one  the 
expansion  will  be  sufficient  to  use  equal  parts  of  this  total  available 
energy.  If  we  compute  the  velocity  for  10  such  sections  we  shall 
have  35  B.  t.  u.  liberated  in  each  one.  By  the  charts  we  may  now 
follow  the  entropy  line  1.71  until  it  intersects  the  desired  total 
heat  line,  and  then  read  the  corresponding  pressure  and  specific 


78 


PROBLEMS 


volume.     By  Table  IV  the  velocity  is  obtained,  and  the  area  of  the 
nozzle  at  any  section  may  then  be  found  from  the  equation 


Area  in  sq.  in.  = 


(Cu.  ft.  flowing  per  sec.)  144 


Velocity  in  ft.  per  sec. 


Hence  the  following  values  may  be  at  once  tabulated : 


Section 

Entropy 

Available 
Energy 
B.  t.  u. 
per  Lb. 

Total 
Heat 
B.  t.  u. 

Pressure 
Lbs.  per 
Sq.  In. 
Abs. 

Specific 
Volume 
Cu.  Ft. 
perLb. 

Velocity 
Ft.  per 
Second 

Area  of 
Nozzle 
Sq.  In. 

Plate 
Used 

Entrance 

1.71 

0 

1,305 

125. 

4.82 

2a 

1 

1.71 

35 

1,270 

89.8 

6.18 

1,323 

0.674 

2a 

2 

1.71 

70 

1,235 

62.8 

8.10 

1,872 

0.624 

2a 

3 

1.71 

105 

1,200 

42.4 

10.9 

2,292 

0.685 

2b 

4 

1.71 

140 

1,165 

27.6 

15.1 

2,647 

0.822 

3a 

5 

1.71 

175 

1,130 

17.1 

22.7 

2,960 

1.101 

3b 

6 

1.71 

210 

1,095 

10.3 

35.4 

3,241 

1.572 

3b 

7 

1.71 

245 

1,060 

6.10 

56.5 

3,501 

2.32 

4b 

8 

1.71 

280 

1,025 

3.44 

94.0 

3,743 

3.62 

4b 

9 

1.71 

315 

990 

1.90 

160. 

3,970 

5.81 

5b 

10 

1.71 

350 

955 

1.00 

285. 

4,185 

9.82 

5b 

If  these  sections  are  now  laid  off  at  equal  intervals  so  that  the 
total  length  becomes  equal  to  the  desired  length  of  the  nozzle  the 
form  will  be  such  that  it  will  give  uniform  acceleration  of  the  steam. 
This  is  theoretically  desirable,  and  is  due  to  the  fact  that  the  sections 
have  been  chosen  at  such  points  that  the  energy  transformed  into 
velocity  in  each  one  is  constant.  Such  a  nozzle  will  have  curved 
elements  for  its  divergent  part  and  will  therefore  be  expensive  to 
make.  It  has  been  found  by  experiment  that  the  divergent  part 
of  the  nozzle  can  be  made  of  straight-line  elements  without  any  great 
loss  in  efficiency.  Since  there  is  a  large  amount  of  hand  work  re- 
quired in  making  nozzles,  they  are  therefore  nearly  always  made 
with  the  divergent  part  as  the  frustum  of  a  pyramid  or  cone.  The 
throat  of  the  nozzle  must  be  approached  by  a  gradually  decreasing 
cross-sectional  area  in  order  to  prevent  loss  due  to  the  sudden  con- 
vergence of  the  stream  lines. 

No  general  rule  can  be  given  to  determine  the  best  length  of 
nozzle  for  all  conditions.  The  length  of  the  rounded  entrance  to  the 


PROBLEMS 


79 


throat  is  a  very  small  portion  of  the  total  length  of  the  nozzle,  and  it 
is  common  practice  for  some  designers  to  have  the  divergent  part 
of  the  nozzle  taper  about  one  in  twelve,  while  others  make  it  only 
about  one  in  twenty. 

44.  For  the  expansion  of  the  previous  problem  find  the  value 
of  the  exponent  n  which  satisfies  the  equation  poV0n  =  p  vn  where  p 
represents  the  pressure  and  v  the  specific  volume  of  the  steam  at  any 
time  during  its  passage  through  the  nozzle  and  pQ  and  v0  refer  to  the 
pressure  and  specific  volume  at  the  entrance  to  nozzle. 

Solution:    Rewriting  the  equation  we  have 


Then  by  the  aid  of  a  log  log  slide  rule,  or  logarithms,  and  the  results 
of  the  previous  problem  the  following  table  may  at  once  be  made: 


Section 

P 

v 

Po 
P 

v 

V0 

n 

Quality  or  Superheat 

Value 

Plate 
Used 

0 

125. 

4.82 

1.000 

1.000 

220° 

2a 

1 

89.8 

6.18 

1.392 

1.283 

1  325 

166° 

2a 

2 

62.8 

8.10 

1.990 

1.680 

1.325 

112° 

2a 

3 

42.4 

10.9 

2.950 

2.265 

1.325 

60° 

2b 

4 

•27.6 

15.1 

4.535 

3.131 

1.325 

5° 

3a 

5 

17.1 

22.7 

7.32 

4.71 

1.285 

97.6% 

3b 

6 

10.3 

35.4 

12.14 

7.35 

1.250 

95.1% 

3b 

7 

6.10 

56.5 

20.50 

11.73 

1.227 

92.6% 

4b 

8 

3.44 

94.0 

36.38 

19.50 

1.210 

90.2% 

4b 

9 

1.90 

160. 

65.80 

33.2 

1.195 

87.9% 

5b 

10 

1.00 

285. 

125.00 

59.1 

1.185 

85.6% 

5b 

This  table  shows  that  so  long  as  the  steam  was  superheated  the 
value  of  n  was  constant,  but  decreased  rapidly  after  reaching  the 
wet  region.  From  Plate  3a,«  it  may  be  seen  that  the  1.71  entropy 
line  crosses  the  saturation  curve  at  a  pressure  of  a  little  more  than 
26  pounds. 


80 


PROBLEMS 


45.  With  the  value  of  n  =  1.325  as  obtained  from  problem 
44,  find  the  pressure  at  the  throat  of  this  nozzle  by  the  following 
equation  :* 

Pt        /     2     \    n 

—  =  \  — T~i  J  n~*  where 
po        Vn  +  I/ 

pt    =  pressure  at  the  throat  of  the  nozzle. 


Po   =  pressure  at  entrance  to  nozzle. 


Solution: 


Pt  = 


/     2    \1325 

=    105  [  -  -  -  )  0.325 

0  V2/ 


2.325/ 

v 

) 


f         I 

=  125  te 

-  125  (im)  =  67-75  lbs- 


=  54.2%  of  po. 

46.  With  the  throat  pressure,  as  determined  from  the  equation 
given  in  the  previous  problem,  determine  the  area  of  the  throat 
and  then  find  by  trial  whether  this  is  the  minimum  section. 

Solution:  Taking  pressures  each  side  of  the  throat  pressure  as 
found  above,  and  obtaining  the  values  needed  as  in  problem  43, 
we  may  construct  the  following  table: 


Section 

Entropy 

Pressure 

Total 
Heat 
B.  t.  u. 

Available 
Energy 
B.  t.  u. 

Velocity 
Ft.  per  Sec. 

Specific 
Volume 

Area 
Sq.  In. 

0 

1.71 

125. 

1,305. 

0 

a 

1.71 

70. 

1,245.8 

59.2 

1,721 

7.46 

0.624 

b 

1.71 

68. 

1,242.9 

62.1 

1,762 

7.62 

0.623 

Throat 

1.71 

67.75 

1,242.5 

62.5 

1,767 

7.64 

0.622 

c 

1.71 

67. 

1,241.5 

63.5 

1,782 

7.70 

0.622 

d 

1.71 

66. 

1,240. 

65. 

1,803 

7.80 

0.623 

*  For  the  derivation  of  this  equation  see  any  good  book  on  turbine  nozzles. 


PROBLEMS 


81 


From  the  above  results,  it  is  seen  that  the  value  of  the  throat 
pressure,  as  found  from  the  equation  of  problem  45,  is  checked  as 
well  as  can  be  desired.  The  computations  from  the  charts  and 
slide  rule  indicate  that  the  pressure  of  67  pounds  would  answer 
equally  well. 

47.  Supposing  that  the  steam  is  compressed,  adiabatically,  from 
the  final  condition  given  in  problem  43,  to  the  pressure  of  125  pounds, 
find  the  value  of  m  to  satisfy  the  equation  pi0  vJJ  =  pv™ 
The  subscript  10  refers  to  the  state  of  the  steam  when  in  the  section 
10  of  problem  43. 


Solution: 

form 


The  equation  may  be  put  in  the  more  convenient 


_p 

Pio 


Section 

P 

V 

P 
PIO 

V13 
V 

m 

Quality  or 
Superheat 
from  44 

10 

1.00 

285. 

1. 

1. 

85.6% 

9 

1.90 

160. 

1.9 

1.783 

.110 

87.9% 

8 

3.44 

94. 

3.44 

3.03 

.115 

90.2% 

7 

6.10 

56.5 

6.10 

5.05 

.117 

92.6% 

6 

10.3 

35.4 

10.3 

8.05 

.118 

95.   % 

5 

17.1 

22.7 

17.1 

12.55 

.122 

96.6% 

4 

27.6 

15.1 

27.6 

18.88 

.128 

5° 

3 

42.4 

10.9 

42.4 

26.12 

.148 

60° 

2 

62.8 

8.10 

62.8 

35.2 

.163 

112° 

1 

89.8 

6.18 

89.8 

46.2 

.175 

166° 

0 

125. 

4.82 

125. 

59.1 

.185 

220° 

From  an  inspection  of  these  exponents  and  those  as  given  by  the 
results  of  44,  it  will  be  seen  that  the  value  of  the  exponent  for 
adiabatic  compression  or  expansion  depends  upon  the  state  of  the 
steam  and  a  simple  general  expression  to  determine  its  value  cannot 
be  given. 


82  PROBLEMS 

When  any  adiabatic  expansion  or  compression  of  steam  is 
plotted  on  the  P-V  diagram  the  resultant  curve  will  always  be 
smooth,  even  though  the  variation  in  the  exponent  may  be  con- 
siderable as  in  the  above  cases.  This  means  that  the  exponents 
change  gradually  from  one  to  the  other  even  when  crossing  the 
saturation  curve.  The  reason  that  the  above  tabulations  do  not 
show  this  more  fully  is  due  to  the  fact  that  the  desire  to  make  a 
short  table  necessitated  the  selection  of  points  which  are  some 
distance  apart. 

48.  In  making  a  turbine  test,  the  barometer  was  read  as  29.55 
inches  at  a  temperature  of  50°  F?  The  diameter  of  the  barometer 
tube  was  0.25  inch  and  the  height  of  the  meniscus  was  0.02 
inch.  The  vacuum  was  determined  by  reading  the  height  of 
mercury  in  a  glass  tube  0.2  inch  in  diameter,  the  bottom  end  of 
which  was  resting  in  a  vessel  of  mercury.  The  height  of  this  mer- 
cury column  was  28.85  inches  at  a  temperature  of  77°  F.  and  the 
meniscus  was  0.04  inch.  The  elevation  of  the  condenser  was 
500  feet  above  sea  level  and  the  barometer  was  in  another  building 
50  feet  higher  than  this.  The  temperature  of  the  atmosphere  -was 
40°  F. 

Find  the  pressure  in  the  condenser  in  inches  of  mercury. 

Solution:    The  barometer  corrections  are: 

(1)  Due  to  capillarity  from  Table  II +  .  012  in. 

(2)  Due  to  temperature  from  Plate  9a +  .024  " 

(3)  Due  to  change  in  elevation  from  Plate  9b +  .055  " 


Total +  .091  in. 

This  third  correction  is  obtained  by  running  along  the  40°  tem- 
perature line  until  we  intersect  the  line  representing  the  average 
altitude,  which  for  this  case  would  be  525  feet,  where  the  correction 
for  100  feet  is  seen  to  be  .110.  Then  for  50-foot  change  in  elevation 
the  correction  would  be  one-half  of  this,  or  .055,  as  above. 

The  corrections  to  the  mercury  column  attached  to  condenser 
are: 


PROBLEMS 


83 


(1)  Due  to  capillarity  from  Table  II +  .  045  in. 

(2)  Due  to  temperature  from  Plate  9a —  .053  " 

Total -  .008  in. 

Hence  by  using  correction  to  nearest  hundredth 
The  barometer  is 

29.55  +  .09  =  29. 64  in.  at  58.1° 
and  the  vacuum  is 

28.85  -  .01  =  28. 84  in.  at  58.1° 
and  absolute  pressure  in  condenser  is 

29.64  -  28.84  =    0.8  in.  Hg. 

NOTE. — Without  making  any  of  the  above  corrections  the  pressure  in  the  condenser  would  appear  to  be 

29.55 28.85  =  0.7  in.  Hg.  and  from  Plate  6b  the  difference  in  total  heats  between  a  pressure  of  0.7 

in.  and  0.8  in.  for  some  common  entropy  line  is  seen  to  be  about  6  or  7  B.  t.  u.  This  amount  of  heat  might 
easily  mean  2%  of  the  total  available  energy  or  as  much  as  15  or  20%  of  the  energy  available  in  the  last 
stage. 


49. 


J 


77 


m     n 


Entropy 
FIG.  8. 

Let  the  area  abcdefka,  Fig.  8,  represent  the  net  work  in  B.  t.  u.  of 
the  theoretical  cycle  for  the  Ferranti  turbine.* 

*  See  Power,  December  30, 1913,  page  908,  for  a  description  of  this  turbine. 


84  PROBLEMS 

If  the  steam  is  received  at  a  pressure  of  145  pounds  per  square 
inch  absolute  and  360  degrees  of  superheat,  then  expands,  adia- 
batically,  to  25  pounds  per  square  inch  absolute,  and,  after  that,  is 
reheated  at  constant  pressure  to  the  initial  temperature  and  then 
finally  expands,  adiabatically,  to  the  back  pressure  of  1  .  5  inches 
of  mercury,  find 

(a)  Net  work  of  the  cycle. 

(b)  Heat  supplied  per  pound  of  steam. 

(c)  Cycle  efficiency. 

(d)  Theoretical  water  rate. 

Solution:  From  Plate  2a,  for  the  pressure  of  145  pounds  and 
superheat  of  360°  we  find 

Total  heat"!    =  Hd   =  1,377  B.  t.  u. 

Jd 


Entropy  1        =1.76 

-jd 


and  from  Plate  3a  for  this  entropy  and  a  pressure  of  25  pounds  we 
have 

Total  heat"!     =  He  =  1,195  B.  t.  u. 

Je 

From  Plates  Ib  and  3b  respectively,  we  also  find  : 
Temperature  of  vaporization  for  145  Ibs.  =  t6  =  356°  F. 
Temperature  of  vaporization  for    25  Ibs.  =  tb,  =  240°  F. 
Hence 

td  =  356  +  360  =  716°  F. 

and  since  the  temperature  after  reheating  is  to  be  the  same  as  this, 
the  superheat  at  /  becomes 

if  -  t,  =  716  -  240  =  476° 

Then,  from  Plate  7,  for  the  pressure  of  25  pounds  and  476°  of 
superheat  we  have 

Total  heat"]     =  H,  =  1386  B.  t.  u. 
J/ 


Entropy]      =  1 . 959  +    or  1 . 96  - 


PROBLEMS  85 

For  this  entropy  and  the  condenser  pressure  of  1.5  inches  of 
mercury,  we  may  then  obtain  from  Plate  5a 


Total  heat!     =  Hfc  =  1077 
-I 


and  from  Plate  5b  the  heat  of  the  liquid  corresponding  to  the 
pressure  of  1  .  5  inches  of  mercury  is 

h0  =  60  B.  t.  u. 
Then  by  inspection  of  Fig.  8  we  have: 

(a)  Net  work  of  cycle  =  area  b'bcdee  +  b'efka 

=  Hd  —  He  +  H/  —  H& 

=  1377  -  1195  +  1386  -  1077 

=  491  B.  t.  u.  per  Ib.  of  steam. 

(b)  Heat  supplied  per  Ib.  of  steam  =  area  labcdefn 

=  Hd  -  He  +  H,  -  ha 

=  1377  -  1195  +  1386  -  60 

=  1508  B.  t.  u. 

491 

(c)  Cycle  efficiency      =  T        =  32.6% 


2545 
(d)  Theoretical  water  rate  is  -r  =  5  .  19  Ibs.  per  h.p.  hr. 


or  -jo-  =  6.94  Ibs.  per  kw.  hr. 


50.  Supposing  the  ordinary  steam  turbine  cycle  abcdj,  Fig. 
8,  had  been  followed,  find  (a)  (b)  (c)  (d)  as  before. 

Solution:  The  only  additional  numerical  value  needed  is  the 
total  heat  at  the  point  j.  From  Plate  5b,  for  the  entropy  1  .  76  and 
the  pressure  of  1  .  5  inches  of  mercury  we  find  : 

Total  heatl  =  H,  =  967.3  B.  t.  u. 
Jj 

Hence 

(a)  Net  work  of  cycle  =  Hd  -  H,-  =  1377  -  967.3  =  409.  7  B.  t.  u. 

(b)  Heat  supplied  per  pound  =  Hd  -  ha  =  1377-60  =  1317  B.  t.  u. 

(c)  Cycle  efficiency  =  y|y  =  31.15% 


86  PROBLEMS 

2545 
(d)  Theoretical  water  rate  =  .^  -  =  6  .  22  Ibs.  per  h.p.  hr. 


3412 
or       40Q  --  =  8.32  Ibs.  per  kw.  hr. 


From  the  above  results  it  is  seen  that  for  the  theoretical  cycles 
the  decrease  in  water  rate  due  to  the  reheating  is 

8-32  -  6-94  _  19  9% 
6.94 

and  the  increase  in  heat  supplied  is 

1508  -  1317        191 

1317  =  13l7  =      4'5% 

or  from  the  cycle  efficiencies  the  gain  due  to  reheating  is 
•326  -  .3115 

,3115  )% 

In  an  actual  turbine  built  to  operate  on  this  cycle  there  would 
be  the  disadvantages  of  higher  first  cost,  more  complications,  and 
greater  radiation  loss.  On  the  other  hand,  by  having  the  steam 
superheated  for  the  entire  passage  through  the  turbine,  or  nearly  so, 
the  leakage,  friction,  and  rotational  losses  are  very  much  reduced. 
Whether  the  reduction  of  these  losses  and  the  slight  increase  in  the 
cycle  efficiency  will  be  sufficient  to  make  this  type  of  turbine  superior 
to  others  cannot  yet  be  told. 


INDEX 


PAGE 

A,  reciprocal  of  the  mechanical  equivalent  of  heat 10 

Absolute  temperature,  °F.,  constant  for,  prob.  37 71 

Abstraction  of  heat  at  constant  volume,  prob.  13 52 

Addition  of  heat  at  constant  pressure,  prob.  9     .     .     .     .     . 48 

Adiabatic  compression,  in  steam  engine,  prob.  35 70 

of  steam,  exponent  for,  prob.  47 81 

Adiabatic  expansion  of  steam,  exponent  for,  probs.  44  and  45    .     .     .     .     .     .     .  79,  80 

F-V  diagram  for,  prob.  12 51 

work  during,  prob.  11 50 

Adiabatics,  definition  of 3 

Air  in  condenser,  prob.  37 71 

Altitude,  variation  of  atmospheric  pressure  with 20 

Area  representing  heat,  general  discussion 3,  5,  9 

supplied  turbine,  prob.  26 61 

Area  representing  work,  general  discussion 1,  2,  4,  5,  7,  10 

Atmosphere,  relative  humidity  of,  prob.  38 72 

Atmospheric  pressure,  the  standard 18 

variation  of,  with  altitude 20 

Available  energy,  definition  of 13 

in  steam  nozzle 17 

Back  pressure,  effect  of,  on  network,  prob.  33 69 

Barometer,  the  thirty-inch  standard 18 

Barometric  corrections : 

due  to  capillarity,  discussion 21 

Table  II 41 

prob.  48 .-._.... 82 

due  to  change  in  elevation,  discussion 20 

Plate  9b 40 

prob.  48 82 

due  to  latitude,  discussion 21 

Table  I 41 

due  to  temperature,  discussion 19 

Plate  9a 40 

probs.  4  and  5 46 

prob.  48 82 

Boiler  and  furnace  efficiency,  prob.  19 57 

Boilers,  determination  of  weight  of  steam  from,  prob.  24       59 

Brass  scales,  effect  of,  on  barometric  corrections 20 

Calorimeter,  throttling,  probs.  15,  16,  17 55,56 

Capillarity,  correction  due  to.     See  Barometric  corrections. 

Complete  expansion  cycle,  prob.  32 68 

87 


88  INDEX 

PAGE 

Compression,  in  uniflow  engine,  prob.  35 70 

of  steam  adiabatically,  prob.  47 81 

Condenser,  determination  of  air  in,  prob.  37 71 

Constant  entropy  changes,  probs.  5,  11,  12,  35,  43,  44,  45,  and  47 46-81 

Constant  pressure  formation  of  steam 6 

Constant  temperature  expansion  of  sup.  steam,  prob.  14 53 

Constant  volume  abstraction  of  heat,  prob.  13 52 

Corrections  of  mercury  volumes.     See  Barometric  corrections. 

Cycle,  Clausius  or  Rankine,  discussion 12 

definition 5 

efficiency,  probs.  26,  27,  29,  49,  and  50 61-85 

Ferranti,  probs.  49  and  50 83,  85 

for  direct-acting  steam-pump,  prob.  34 69 

volumes  for  complete  and  incomplete  exp.,  prob.  32 68 

Dalton's  law,  application  of,  prob.  38 72 

Degrees  of  superheat,  definition    . ;    9^ 

Delivered  thermal  efficiency,  probs.  26  and  27     .     .     . 61,  63 

Density  of  mercury,  discussion 22 

Table  III 41 

Direct-acting  steam-pump  cycle,  prob.  34 69 

Efficiency,  cycle,  probs.  26  and  27 61,  63 

delivered  thermal,  probs.  26  and  27 61,  63 

of  boiler  and  furnace,  prob.  19 57 

of  Ferranti  cycle,  probs.  49  and  50 83,  85 

Ratio,  probs.  26  and  27 61,  63 

Energy,  available,  definition 13 

Intrinsic,  definition 5,  11 

how  found,  probs.  8  and  11 48,50 

of  water,  prob.  29 64 

Entropies,  constant,  how  drawn 15 

probs.  5,  11,  12,  35,  43-45,  and  47 46-81 

Entropy,  definition 2 

of  steam,  definition .12 

how  found  from  chart,  probs.  1  and  2 45 

Equivalent  evaporation,  probs.  20  and  21 57,  58 

Exhaust  opening  in  large  turbine,  prob.  40 .74 

Exhaust,  steam  for  heating,  prob.  25 61 

Expansion  of  steam,  adiabatically,  exponent  for,  probs.  44  and  45 79,  80 

in  single  stage  turbine,  prob.  28 64 

work  during  adiabatic,  prob.  11 50 

Expansion,  ratio  of,  definition,  prob.  11 50 

Exponent  for  adiabatic  compression  of  steam,  prob.  47 81 

Exponent  for  adiabatic  expansion  of  steam,  prob.  44  and  45 79,  80 

External  latent  heat,  definition 10 

External  work  at  constant  pressure,  discussion 9,  10 

Plates  8a  and  8b 38,39 

problem  7 47 

Factor  of  evaporation,  probs.  20  and  21 57,  58 


INDEX  89 

PAGE 

Ferranti  turbine,  probs.  49  and  50 83,  85 

Flow  of  steam  through  nozzles,  probs.  41  and  42 74,  75 

Free  expansion 3 

Friction  in  steam-pipe,  prob.  23 59 

General  equation  for  work  and  heat 5 

Gravity,  reduction  to  standard 21 

Heat  abstracted  at  constant  volume,  prob.  13 52 

Heat  added,  at  constant  pressure,  probs.  9  and  10        48,  49 

at  constant  temperature  to  super,  steam,  prob.  14 53 

Heating  by  exhaust  steam,  prob.  25 61 

Heat,  intrinsic,  definition , 5,  11 

how  found,  probs.  8  and  11 48,  50 

of  water,  prob.  29      .......    V    ......(,......  64 

Heat  of  the  liquid  curve,  how  drawn 15 

definition 7 

Heat  of  the  liquid,  how  found,  prob.  2 45 

Heat  of  superheat,  definition 9 

Heat  Volume  Chart,  Plates  1  to  7 24-36 

Humidity,  relative,  and  vapor  pressure,  prob.  38 72 

Incomplete  expansion,  probs.  29,  80,  and  31 64-68 

Index  plate  for  main  chart 23 

Intrinsic  heat,  definition 5,  11 

how  obtained,  probs.  8  and  11 48,50 

of  water,  prob.  29 64 

Irreversible  adiabatics,  definition 3,  4 

Isothermal  expansion  of  superheated  steam,  prob.  14 53 

Mechanical  equivalent  of  heat 10 

Mercury  column,  corrections  for.     See  Barometric  corrections. 

Mercury,  density  of,  discussion 22 

Table  III 41 

Mixture  of  wet  and  superheated  steam,  prob.  22 58 

Net  work  of  cycle,  how  affected  by  back  pressure,  prob.  33 69 

Net  work  of  incomplete  expansion  cycle,  two  methods  of  solving,  probs.  29  and  30.  64,  67 

Non-expansion  cycle,  prob.  34 69 

Nozzle,  size  of,  for  given  flow,  probs.  41  and  42 74,  75 

throat  pressures,  probs.  41,  45,  and  46 74-80 

velocities,  probs.  28,  41-43 64,  74-77 

Table  IV       42,43 

Nozzles,  divergent,  prob.  43 77 

non-divergent,  prob.  41 74 

Pipe,  size  of,  for  turbine,  prob.  39 73 

Plates  1-7,  Heat- Volume  charts 24-36 

8,  External  Work- Volume  chart 38,  39 

9a  and  9b,  Corrections  to  Barometer 40 

Preparation  and  use  of  the  charts  and  table  of  velocities 14-17 

Pressure,  effect  of,  excessive  drop  in  steam  main,  prob.  23 59 

how  found,  from  quality  and  specific  volume,  prob.  3 45 

P-V  diagram,  discussion       1 


90  INDEX 

PAGE 

P-V  diagram  for  steam-engine  cycle,  prob.  29 64 

of  adiabatic  expansion,  prob.  12 51 

Properties,  independent  of  path 5 

Properties  of  steam g_12 

Quality,  at  end  of  adiabatic  expansion,  prob.  11 50 

at  end  of  constant  volume  change,  prob.  13 .52 

at  release,  prob.  36 71 

definition  of     .     .     .     . 8 

line  of  constant,  how  drawn 15 

obtained  by  throttling  calorimeter,  probs.  15-17 55,  56 

Ratio  of  expansion,  definition,  prob.  11 50 

Ratio  of  pressures  in  nozzles,  probs.  41  and  45 74,  80 

Ratio  of  volumes  for  complete  and  incomplete  expansion  cycles,  prob.  32    .     .     .     .68 

Reduction  of  barometric  readings  to  standard  gravity      . 21 

Reduction  of  mercury  column  to  pounds  per  square  inch 22 

Reheating,  effect  of,  in  multi-stage  turbine,  prob.  4® 75 

Relative  humidity  and  vapor  pressure,  prob.  38  .     . 72 

Reversible  adiabatics,  definition 3,  4 

Saturation  curve,  definition 7 

Scales,  selection  of,  for  main  chart 14,  15 

Single-stage  turbine,  prob.  28 64 

Size  of  turbine  exhaust  pipe,  prob.  40 74 

supply  pipe,  prob.  39 73 

Specific  heat  of  superheated  steam  for  very  low  pressure,  how  found 14 

Specific  volume  of  steam,  definition 11 

of  water,  numerical  values 11 

Standard  atmospheric  pressure,  definition  . 18 

Steam,  properties  of 6-12 

superheated,  definition 8 

Superheat  at  release,  prob.  36  .. 71 

Superheated  steam  and  wet  steam  mixed,  prob.  22 58 

Superheated  steam  in  the  atmosphere,  prob.  38 72 

Superheating  due  to  throttling,  probs.  15-18 55,  56 

Tables  I,  II,  III 41 

Table  IV    .     . .     42-43 

Table  of  velocities,  how  prepared 17 

Temperature,  absolute,  °F.,  prob.  37      .     . 71 

Temperature  correction  of  barometer 19 

Temperature-entropy  diagram,  discussion        1 

for  Ferranti  cycle,  prob.  49 83 

for  turbine  cycle,  prob.  26 61 

Temperature  of  superheated  steam  for  very  low  pressures,  scale  of 15 

Temperature  of  vaporization,  curve  of .16 

Temperatures,  how  found  from  chart,  probs.  1  and  2 45 

use  of  scale  of  approximate,  prob.  38   .     .     . 72 

Thermal  efficiency,  delivered,  probs.  26  and  27 61,  63 

Throat  pressures  in  nozzles,  probs.  41,  4$,  4® 74,  80 

Throttling  calorimeter,  probs.  15-17 55,  56 


INDEX  91 

PAGE 

Throttling  in  steam  automobile,  prob.  18 56 

Total  entropy  of  steam 12 

Total  heat,  discussion 7-9 

Turbine,  Ferranti,  probs.  49  and  50 83,  85 

multi-stage,  probs.  41  and  42 74,  75 

single-stage,  prob.  28 64 

Uniflow  engine,  compression  in,  prob.  35 70 

Using  the  charts,  general  method  of 17 

Vacuum,  effect  of,  on  water  rate,  prob.  27 63 

Velocities,  jet,  Table  IV  42,  43 

how  computed 16 

Velocity  of  steam,  at  exhaust,  from  turbine,  prob.  40 74 

through  nozzle,  probs.  28,  41-43 64,  74-77 

through  pipe  to  turbine,  prob.  39 73 

Volume,  increase  of,  during  superheating,  probs.  9  and  10 48,  49 

Volume  of  water  on  P-V  diagram,  prob.  29 64 

Volume,  specific,  definition 11 

how  found,  from  chart,  probs.  1  and  2 45 

Volumes  for  complete  and  incomplete  expansion,  prob.  32 68 

Water,  specific  volume  of 7,  11 

prob.  29  s 64 

Water  rate  and  vacuum,  prob.  27 63 

Water  rate  of  large  turbines,  probs.  26,  27,  and  40 61-63,  74 

of  Ferranti  turbine,  theoretical,  probs.  49  and  50 83,  85 

Weight  of  steam  from  boilers,  how  computed,  prob.  24 59 

Weight  of  steam  in  pipe,  how  computed,  prob.  6 47 

Wet  steam,  total  heat  of 8 

Work  done  during  adiabatic  expansion,  prob.  11 50 

during  constant  pressure  expansion,  discussion, 9,  10 

prob.  7 47 

Work,  general  equation  for  .  .1 

lost  due  to  incomplete  expansion,  prob.  31 68 


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